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previous year question of Differential Equations for NDA

Previous Year Differential Equation Questions For NDA

Previous Year Differential Equation Questions
Q 2713791640

What is `(d^2 x)/(dy^2)` equal to?
NDA Paper 1 2017
(A)

` - ((d^2 y)/(dx^2))^(-1) ((dy)/(dx))^(-3)`

(B)

`((d^2 y)/(dx^2))^(-1) ((dy)/(dx))^(-2)`

(C)

`- ((d^2 y)/(dx^2)) ((dy)/(dx))^(-3)`

(D)

`((d^2 y)/(dx^2))^(-1)`

Solution:

`(d^2x)/(dy^2) = d/(dy) ((dx)/(dy)) = d/(dy) ((dy)/(dx))^(-1)`

`=- ((dy)/(dx))^(-2) (d^2y)/(dy dx)`

`=-((dy)/(dx))^(-2) * (d^2 y)/(dx^2) ((dy)/(dx))^(-1)`

`=-((dy)/(dx))^(-3) (d^2 y)/(dx^2)`
Correct Answer is `=>` (C) `- ((d^2 y)/(dx^2)) ((dy)/(dx))^(-3)`
Q 2723780641

If `x dy = y{dx + ydy); y(1) = 1` and `y(x) > 0 ,` then what is `y(- 3)` equal to?
NDA Paper 1 2017
(A)

`3`

(B)

`2`

(C)

`1`

(D)

`0`

Solution:

`xdy =ydx + y^2 dy`

`int (xdy -ydx)/(y^2) = int dy`

`- int d (x/y) = y+ C`

`- x/y = y+ C`

`y(1)=1`

`-1 = 1+ C`

`=> (-x)/y = y-2`

`y^2 - 2y + x =0`

`(y-1)^2 = 1-x`

`y= 1+ sqrt (1-x)`

`y(-3) = 1+ sqrt 4 =3`
Correct Answer is `=>` (A) `3`
Q 2733880742

What are the degree and order respectively of the differential equation

`y= x ((dy)/(dx))^2 +((dx)/(dy))^2 ` ?
NDA Paper 1 2017
(A)

`1,2`

(B)

`2,1`

(C)

`1,4`

(D)

`4,1`

Solution:

`y= x ((dy)/(dx))^2 + (1/((dy)/(dx))^2)`

`4((dy)/(dx))^2 = x ((dy)/(dx))^4 +1`

Order `=1`

Degree `=4`
Correct Answer is `=>` (D) `4,1`
Q 2703880748

What is the differential equation corresponding to `y^2 -2ay+ x^ 2 = a^ 2` by eliminating `a`?

where, `p= (dy)/(dx)`
NDA Paper 1 2017
(A)

`(x^2 -2y^2) p^2-4pxy-x^2=0`

(B)

`(x^2-2y^2)p^2 + 4 pxy -x^2=0`

(C)

`(x^2+ 2y^2) p^2 - 4pxy-x^2=0`

(D)

`(x^2+ 2y^2) p^2 - 4 pxy+ x^2=0`

Solution:

`y^2 - 2ay +x^2 =a^2`

`2y (dy)/(dx) -(2a dy)/(dx) + 2x=0`

`a = (y+x)/((dy)/(dx))`

on putting the value of `a`

`y^2 - 2ay +x^2 =a^2`

`y^2 -2y (y+x)/((dy)/(dx))+x^2 =(y+x)/((dy)/(dx))^2`

`(x^2 -2y^2) ((dy)/(dx))^2 + (dy)/(dx) xy -x^2 =0`

`(x^2 + 2y^2) p^2 + 4pxy -x^2=0`
Correct Answer is `=>` (B) `(x^2-2y^2)p^2 + 4 pxy -x^2=0`
Q 2753080844

What is the general solution of the differential equation

`ydx-(x+2y^2 )dy=0` ?
NDA Paper 1 2017
(A)

`x=y^2 + cy`

(B)

`x= 2cy^2`

(C)

`x= 2y^2 + cy`

(D)

None of the above

Solution:

`y dx -(x + 2y^2) dy =0`

`int (ydx - xdy)/(y^2) = int 2 dy`

`int d (x/y) = 2y + c`

`x= 2y^2 + cy`
Correct Answer is `=>` (C) `x= 2y^2 + cy`
Q 2713191049

What is the solution of the differential equation `ln ((dy)/(dx)) -a=0` ?
NDA Paper 1 2017
(A)

`y=xe^a + c`

(B)

`x=ye^a + c`

(C)

`y=lnx+c`

(D)

`x=lny+c`

Solution:

` ln ((dy)/(dx)) - a =0`

`(dy)/(dx) = e^a`

`int dy = int e^a dx`

`y= x e^a + C`
Correct Answer is `=>` (A) `y=xe^a + c`
Q 2136791672

What are the order and degree respectively of the
differential equation whose solution is
`y =cx+ c ^(2) - 3c^( 3//2) + 2`, where `c` is a parameter?
NDA Paper 1 2016
(A)

`1, 2`

(B)

`2, 2`

(C)

`1, 3`

(D)

`1, 4`

Solution:

Given, `y =cx+ c^(2) - 3c^(3//2 ) + 2`...........(i)

On differentiating both sides w.r.t. x, we get

`dy/dx = c ` ..............(ii)

From Eqs. (i) and (ii), we have

`y = dy/dx xx x + (dy/dx)^(2) -3(dy/dx)^(3//2) +2`

`=> y- x dy/dx -(dy/dx)^(2) -2 = -3 (dy/dx)^(3//2) `

`=> [y - x (dy/dx ) -(dy/dx)^(2) -2]^(2) = 9 (dy/dx)^(3)`

Hence, order is `1` and degree is `4`.
Correct Answer is `=>` (D) `1, 4`
Q 1609834718

What is the solution of the differential equation

`(ydx - xdy)/y^2 = 0`?

where, `C` is an arbitrary constant.
NDA Paper 1 2015
(A)

`xy =C`

(B)

`y =Cx`

(C)

`x+y = C`

(D)

`x- y =C`

Solution:

Consider the given differential equation,

`(ydx - xdy)/y^2 = 0`

` => d (x/y) = 0 quad ( :. d(u/v)= (v . du - u.dv)/v^2 )`

On integrating both side, we get

`=> int d ( x/y) = C_1 => x/y = C_1 => x = C_1Y`

`=> y = 1/C_1 x => y = Cx`, where `C = 1/C_1`.
Correct Answer is `=>` (B) `y =Cx`
Q 1659034814

What is the solution of the differential equation

`sin((dy)/(dx)) -a = 0` ?

where `C` is an arbitrary constant.
NDA Paper 1 2015
(A)

`y =xsin^(-1) a+ C`

(B)

`x =y sin^(-1) a+ C`

(C)

`y = x + xsin^(-1) a+ C`

(D)

`y = sin^(-1) a+ C`

Solution:

Consider the given differential equation,

`sin((dy)/(dx)) -a = 0`

` => sin((dy)/(dx)) = a => (dy)/(dx) = sin^(-1) a`

` => dy = (sin^(-1) a) dx`

On integrating both side w.r.t x, we get

` int dy = int (sin^(-1) a) dx => y = (sin^(-1) a) int dx`

`=> y = (sin^(-1) a)- x + C`
Correct Answer is `=>` (A) `y =xsin^(-1) a+ C`
Q 1619034819

What is the solution of the differential equation

`(dx)/(dy) + x/y - y^2 = 0?`

where, `C` is an arbitrary constant.
NDA Paper 1 2015
(A)

`xy = x^4 + C`

(B)

`xy = y^4 + C`

(C)

`4xy = y^4 + C`

(D)

`3xy = y^3 + C`

Solution:

Consider the given differential equation,

`(dx)/(dy) + x/y - y^2 `

Which is linear differential equation of the form

` (dx)/(dy) + Px = Q`, where, `P` and `Q` are constant or function of

`y` only

Here, `P = 1/y` and `Q = y^2`

`:. I.F. = e ^(int Pdy) =e^(int 1/y dy) = e^(int In (y)) = y`

Now, the solution of given differential equation is

` x. y = int (y. y^2 ) dy + C_1`

`=> x . y = int y^3 dy + C_1 => x. y = y^4/4 + C_1`

`=> 4xy = y^4 + C_1`, where `C = 4C_1`
Correct Answer is `=>` (C) `4xy = y^4 + C`
Q 2281691527

The degree of the differential equation

` (dy)/(dx) - x = ( y - x (dy)/(dx))^(-4)` is
NDA Paper 1 2015
(A)

`2`

(B)

`3`

(C)

`4`

(D)

`5`

Solution:

Given , ` (dy)/(dx) - x = ( y - x (dy)/(dx))^(-4)`

` => ( y - x (dy)/(dx))^(4) ((dy)/(dx) - x ) = 1`

Hence, the degree of differential equation is `5`.
Correct Answer is `=>` (D) `5`
Q 2231891722

The solution of ` (dy)/(dx) = sqrt(1 - x^2 - y^2 + x^2 y^2 )` is
NDA Paper 1 2015
(A)

`sin^(-1) y = sin^(-1) x + C`

(B)

`2sin^(-1) y = sqrt(1- x^2) + sin^(-1) x + C`

(C)

`2sin^(-1) y = xsqrt(1- x^2) + sin^(-1) x + C`

(D)

`2sin^(-1) y = x sqrt(1- x^2) + cos^(-1) x + C`

Solution:

Given, ` (dy)/(dx) = sqrt(1 - x^2 - y^2 + x^2 y^2 )`

` = sqrt((1 - x^2) - y^2 (1 - x^2) )`

` = sqrt((1 - x^2) (1 - y^2) ) = ( sqrt(1 - x^2)) ( sqrt(1 - y^2)) `

` => (dy)/ sqrt(1 - y^2) = sqrt(1 - x^2) dx`

On integrating both sides, we get

` int (dy)/ sqrt(1 - y^2) = int sqrt(1 - x^2)dx `

` => sin ^(-1) y = 1/2 [ x sqrt(1 - x^2) + sin ^(-1) x] +A`

` =>2 sin ^(-1) y = x sqrt(1 - x^2) + sin ^(-1) x + C quad \ \ \ \ \ \[∵ C = 2A]`

`"Alternatively :"`

Check by differentiating option

Differentiating option 3

`2 sin^(-1) y= x sqrt(1-x^2) + sin^(-1) x + c`

`2 1/( sqrt(1- y^2)) (dy)/(dx) = (x (1) * (-2x))/(2 sqrt(1-x^2)) + sqrt(1-x^2) + 1/( sqrt(1-x^2))`

`2 1/(sqrt(1- y^2)) (dy)/(dx) =(-x^2)/( sqrt(1-x^2)) + 1/(sqrt(1-x^2)) + sqrt(1-x^2)`

`2 1/(sqrt(1- y^2)) (dy)/(dx) = (-x^2 + 1 + 1-x^2)/( sqrt(1-x^2)) =(2 (1-x^2))/( sqrt(1-x^2)) = 2sqrt(1-x^2)`

`(dy)/(dx) = sqrt(1-x^2 -y^2- x^2y^2)`
Correct Answer is `=>` (C) `2sin^(-1) y = xsqrt(1- x^2) + sin^(-1) x + C`
Q 2221091821

The differential equation of the family of circles passing
through the origin and having centres on the `X`-axis is
NDA Paper 1 2015
(A)

`2xy (dy)/(dx) = x^2 - y^2`

(B)

`2xy (dy)/(dx)= y^2 - x^2`

(C)

` 2xy (dy)/(dx) = x^2 + y^2`

(D)

` 2xy (dy)/(dx) = x^2 + y^2 = 0`

Solution:

Let equation of family of circles passing through origin

and having centre `(a, 0)` be

`(x- a)^2 + (y- 0)^2 = a^ 2`

`=> x^2 + a^2 -2ax+ y^2 =a^2`

` => x^2 + y^2 - 2ax = 0` ..........(i)

On differentiating Eq. (i), we get

` => 2x+ 2y (dy)/(dx) - 2a =0`

` => x+ y (dy)/(dx) - a =0`

` => x + y (dy)/(dx) - [(x^2 + y^2)/(2x) ] = 0`

` => 2x^2 + 2xy (dy)/(dx) -x^2 - y^2 = 0`

`=> 2xy (dy)/(dx) + x^2 - y^2 = 0`

` => 2xy (dy)/(dx) = y^2 - x^2`
Correct Answer is `=>` (B) `2xy (dy)/(dx)= y^2 - x^2`
Q 2211191920

The order and degree of the differential equation of
parabolas having vertex at the origin and focus at `(a, 0)`,
where `a > 0`, are respectively
NDA Paper 1 2015
(A)

`1,1`

(B)

`2, 1`

(C)

`1, 2`

(D)

`2, 2`

Solution:

Equation of parabola having vertex at the origin and

focus at `(a, 0)`, where `a > 0`, is `y^2 = 4ax`

On differentiating the above equation, we get

`2y (dy)/(dx) = 4a`

`=> y(dy)/(dx) = 2a`

`=> y (d^2y)/(dx^2) + ((dy)/(dx))^2 = 0`

Hence, the order and degree of differential equation

are `2` and `1`, respectively.
Correct Answer is `=>` (B) `2, 1`
Q 1649545413

Which of the following statements is/are correct in
respect of regression coefficients?

I. It measures the degree of linear relationship
between two variables.

II. It gives the value by which one variable changes
for a unit change in the other variable.

Select the correct answer using the code given
below.
NDA Paper 1 2015
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Hence, both statements are true.
Correct Answer is `=>` (C) Both I and II
Q 2374056856

What is the general solution of the differential equation `xdy - y dx = y^2 ?`
NDA Paper 1 2014
(A)

`x = Cy`

(B)

`y^2 = Cx`

(C)

`x + xy -Cy =0`

(D)

None of these

Solution:

Given differential equation,
`xdy- ydx = y^2`

`=> - (ydx - xdy) = y^2`

`=> - ((ydx - x dy)/y^2) =1`

`=> -d(x/y) =1`

`=> d(x/y)=1`

On integrating both sides, we get

`x/y = C => x =Cy`

which is the required general solution.
`
Correct Answer is `=>` (A) `x = Cy`
Q 1762891735

The general solution of the
differential equation `( x^2 + x + 1) dy + (y^ 2 + y + 1) dx = 0` is
`(x + y + 1) = A(1 + Bx + Cy + Dxy)`, where `B , C` and `D` are
constants and `A` is parameter.

What is `B` equal to?
NDA Paper 1 2014
(A)

`-1`

(B)

`1`

(C)

`2`

(D)

None of these

Solution:

Given differential equation is

`(x^2 + x + 1)dy + (y^2 + y + 1)dx = 0`

` =>(x^2 + x + 1)dy = - (y^2 + y + 1)dx `

`=> (dy)/( 1 + y + y^2) = - (dx)/( 1 + x + x^2)`

` => (dx)/( 1 + x + x^2) + (dy)/( 1 + y + y^2) = 0`

` => int (dx)/((x + 1/2)^2 + (3^2)/4) + int (dy)/((y + 1/2)^2 + (3)/4 ) = 0`

` => int (dx)/((x + 1/2)^2 + (sqrt(3)/2)^2) + int (dy)/((y + 1/2)^2 + (sqrt(3)/2)^2) = 0`

` => 1/(sqrt(3)/2) tan^(-1) {(x + 1/2)/sqrt(3)/2} + 1/(sqrt(3)/2 ) tan^(-1) { (y + 1/2)/sqrt(3)/2}`

` = 2/sqrt(3) tan^(-1) C_1 quad ( ∵ int (dx)/(a^2 + x^2) = 1/a tan^(-1) x/a)`

` => 2/sqrt(3) tan^(-1) ( (2x+1)/sqrt(3)) + 2/sqrt(3) tan^(-1)( (2y +1)/sqrt(3)) = 2/sqrt(3) tan^(-1) C_1`

`=> tan^(-1) (2x+1)/sqrt(3) + tan^(-1) (2y +1)/sqrt(3) = tan^(-1) C_1`

` => tan^(-1) { ( (2x+1)/sqrt(3) + (2y +1)/sqrt(3) )/( 1 - (2x+1)/sqrt(3) . (2y +1)/sqrt(3) )} = tan^(-1) C_1`

`[ ∵ tan^(-1) x + tan^(-1) y = tan^(-1) ( (x +y) /(1 + xy) ) ]`

` => (sqrt(3) [(2x + 1) + (2y + 1)])/(3- (2x + 1). (2y + 1) ) = C_1 `

` => (sqrt(3) (2x + 2y + 2) ) /( 3 - ( 4xy + 2 y + 2x + 1) ) = C_1 `

` => (2 sqrt (3) (x + y + 1))/( - 4xy -2y -2x + 2) = C_1`

` => 2 sqrt (3) (x + y + 1) = C_1 (2- 2x- 2y- 4xy)`

` => 2 sqrt (3) (x + y + 1) = 2 C_1(1- x- y- 2xy)`

` => (x + y + 1) = C_1 /sqrt(3) (1- x- y- 2xy)`

On comparing with

` (x + y + 1) = A (1 + Bx + Cy + Dxy)`

Here, `A` is parameter and `B, C` and `D` are constants.

The value of `B = -1`
Correct Answer is `=>` (A) `-1`
Q 1732091832

The general solution of the
differential equation `( x^2 + x + 1) dy + (y^ 2 + y + 1) dx = 0` is
`(x + y + 1) = A(1 + Bx + Cy + Dxy)`, where `B , C` and `D` are
constants and `A` is parameter.

What is `C` equal to?
NDA Paper 1 2014
(A)

`1`

(B)

`-1`

(C)

`2`

(D)

None of these

Solution:

Given differential equation is

`(x^2 + x + 1)dy + (y^2 + y + 1)dx = 0`

` =>(x^2 + x + 1)dy = - (y^2 + y + 1)dx `

`=> (dy)/( 1 + y + y^2) = - (dx)/( 1 + x + x^2)`

` => (dx)/( 1 + x + x^2) + (dy)/( 1 + y + y^2) = 0`

` => int (dx)/((x + 1/2)^2 + (3^2)/4) + int (dy)/((y + 1/2)^2 + (3)/4 ) = 0`

` => int (dx)/((x + 1/2)^2 + (sqrt(3)/2)^2) + int (dy)/((y + 1/2)^2 + (sqrt(3)/2)^2) = 0`

` => 1/(sqrt(3)/2) tan^(-1) {(x + 1/2)/sqrt(3)/2} + 1/(sqrt(3)/2 ) tan^(-1) { (y + 1/2)/sqrt(3)/2}`

` = 2/sqrt(3) tan^(-1) C_1 quad ( ∵ int (dx)/(a^2 + x^2) = 1/a tan^(-1) x/a)`

` => 2/sqrt(3) tan^(-1) ( (2x+1)/sqrt(3)) + 2/sqrt(3) tan^(-1)( (2y +1)/sqrt(3)) = 2/sqrt(3) tan^(-1) C_1`

`=> tan^(-1) (2x+1)/sqrt(3) + tan^(-1) (2y +1)/sqrt(3) = tan^(-1) C_1`

` => tan^(-1) { ( (2x+1)/sqrt(3) + (2y +1)/sqrt(3) )/( 1 - (2x+1)/sqrt(3) . (2y +1)/sqrt(3) )} = tan^(-1) C_1`

`[ ∵ tan^(-1) x + tan^(-1) y = tan^(-1) ( (x +y) /(1 + xy) ) ]`

` => (sqrtr(3) [(2x + 1) + (2y + 1)])/(3- (2x + 1). (2y + 1) ) = C_1 `

` => (sqrtr(3) (2x + 2y + 2) ) /( 3 - ( 4xy + 2 y + 2x + 1) ) = C_1 `

` => (2 sqrt (3) (x + y + 1))/( - 4xy -2y -2x + 2) = C_1`

` => 2 sqrt (3) (x + y + 1) = C_1 (2- 2x- 2y- 4xy)`

` => 2 sqrt (3) (x + y + 1) = 2 C_1(1- x- y- 2xy)`

` => (x + y + 1) = C_1 /sqrt(3) (1- x- y- 2xy)`

On comparing with

` (x + y + 1) = A (1 + Bx + Cy + Dxy)`

Here, `A` is parameter and `B, C` and `D` are constants.

The value of `C = -1`
Correct Answer is `=>` (B) `-1`
Q 1762091835

The general solution of the
differential equation `( x^2 + x + 1) dy + (y^ 2 + y + 1) dx = 0` is
`(x + y + 1) = A(1 + Bx + Cy + Dxy)`, where `B , C` and `D` are
constants and `A` is parameter.

What is `D` equal to?
NDA Paper 1 2014
(A)

`-1`

(B)

`1`

(C)

`-2`

(D)

None of these

Solution:

Given differential equation is

`(x^2 + x + 1)dy + (y^2 + y + 1)dx = 0`

` =>(x^2 + x + 1)dy = - (y^2 + y + 1)dx `

`=> (dy)/( 1 + y + y^2) = - (dx)/( 1 + x + x^2)`

` => (dx)/( 1 + x + x^2) + (dy)/( 1 + y + y^2) = 0`

` => int (dx)/((x + 1/2)^2 + (3^2)/4) + int (dy)/((y + 1/2)^2 + (3)/4 ) = 0`

` => int (dx)/((x + 1/2)^2 + (sqrt(3)/2)^2) + int (dy)/((y + 1/2)^2 + (sqrt(3)/2)^2) = 0`

` => 1/(sqrt(3)/2) tan^(-1) {(x + 1/2)/sqrt(3)/2} + 1/(sqrt(3)/2 ) tan^(-1) { (y + 1/2)/sqrt(3)/2}`

` = 2/sqrt(3) tan^(-1) C_1 quad ( ∵ int (dx)/(a^2 + x^2) = 1/a tan^(-1) x/a)`

` => 2/sqrt(3) tan^(-1) ( (2x+1)/sqrt(3)) + 2/sqrt(3) tan^(-1)( (2y +1)/sqrt(3)) = 2/sqrt(3) tan^(-1) C_1`

`=> tan^(-1) (2x+1)/sqrt(3) + tan^(-1) (2y +1)/sqrt(3) = tan^(-1) C_1`

` => tan^(-1) { ( (2x+1)/sqrt(3) + (2y +1)/sqrt(3) )/( 1 - (2x+1)/sqrt(3) . (2y +1)/sqrt(3) )} = tan^(-1) C_1`

`[ ∵ tan^(-1) x + tan^(-1) y = tan^(-1) ( (x +y) /(1 + xy) ) ]`

` => (sqrtr(3) [(2x + 1) + (2y + 1)])/(3- (2x + 1). (2y + 1) ) = C_1 `

` => (sqrtr(3) (2x + 2y + 2) ) /( 3 - ( 4xy + 2 y + 2x + 1) ) = C_1 `

` => (2 sqrt (3) (x + y + 1))/( - 4xy -2y -2x + 2) = C_1`

` => 2 sqrt (3) (x + y + 1) = C_1 (2- 2x- 2y- 4xy)`

` => 2 sqrt (3) (x + y + 1) = 2 C_1(1- x- y- 2xy)`

` => (x + y + 1) = C_1 /sqrt(3) (1- x- y- 2xy)`

On comparing with

` (x + y + 1) = A (1 + Bx + Cy + Dxy)`

Here, `A` is parameter and `B, C` and `D` are constants.

The value of `D = -2`
Correct Answer is `=>` (C) `-2`
Q 1763812745

The solution of ` (dy)/(dx) = |x|` is

where, `C` is an arbitrary constant.

NDA Paper 1 2014
(A)

` y = (x |x|)/2 + C`

(B)

` y = ( |x|)/2 + C`

(C)

` y = x^2/2 + C`

(D)

` y = x^3/2 + C`

Solution:

Given differential equation,

` (dx)/(dy) = |x|`

` dy = |x| dx` ....(1)

Case 1. if ` x > 0`

` int dy = int xdx`(on integrating)

` => y = x^2/2 + C => y = (x.(x))/2 + C` ........(2)

Case 2. if ` x < 0`

` int dy = - int xdx`(on integrating)

` => y = (-x^2)/2 + C => y = (x.(x))/2 + C` .......(3)

When we combined both cases, we get the required

solution.

` y = (x|x|)/2 + C`
Correct Answer is `=>` (A) ` y = (x |x|)/2 + C`
Q 1763012845

What is the solution of `((dy)/(dx)) + 2y = 1` satisfying `y(0) = 0?`
NDA Paper 1 2014
(A)

` y = (1-e^(-2x))/2`

(B)

` y = (1+e^(-2x))/2`

(C)

`y = 1 + e^x`

(D)

`y = (1 + e^x)/2`

Solution:

Given differential equation

`(dy)/(dx) + 2y = 1`

On comparing with `(dy)/(dx) + Py = Q` we get

`P = 2` and `Q =1`

Now, `IF = e ^(int pdx) = e ^(int 2dx) = e^(2x)`

Then, complete solution

`y . = int Q(IF)dx + C = int 1.e ^(2x) dx + C`

` => y.e^(2x) = 1/2 e^(2x) + C`

` => y = 1/2 + C . e^(-2x)` ..........(1)

Now, at` x = 0 => y = 0`

From Eq. (1),

` 0 = 1/2 + C.e^0 => 1/2 + C = 0 => C=- 1/2`

:. Required solution is

` y = 1/2 - 1/2 . e^(-2x) => y = (1-e^(-2x))/2`
Correct Answer is `=>` (A) ` y = (1-e^(-2x))/2`
Q 1713456349

What is the number of arbitrary constants in the
particular solution of differential equation of third
order?
NDA Paper 1 2014
(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

We know that, the solution obtained by giving

particular values to the arbitrary constants in the general

solution is called a particular solution. i.e., In particular

solution of differential of any order, there is no arbitrary

constants are present.

Hence, the number of arbitrary constants in the particular

solution of differential equation of third order is zero.
Correct Answer is `=>` (A) `0`
Q 1713556440

What is the equation of a curve passing through `(0, 1)`
and whose differential equation is given by
`dy = y tan x dx?`
NDA Paper 1 2014
(A)

`y = cos x`

(B)

`y = sin x`

(C)

`y = sec x`

(D)

`y = cosec x`

Solution:

The given differential equation of the curve is,

` dy = y tan x . dx`

` => int (dy)/y = int tan x . dx` (on integrating)

` => log y = log sec x + log C`

`=> log y = log C . sec x`

`=> y = C . sec x` ..........(1)

Since, the curve passes through the origin `(0, 1 )`, then

` 1 = C . sec 0 => C = 1`

`:.` Required equation of curve is,

` y = sec x`
Correct Answer is `=>` (C) `y = sec x`
Q 1723556441

Consider the following statements in respect of the
differential equation `(d^2y)/(dx^2) + cos (( dy)/(dx)) = 0`

1. The degree of the differential equation is not defined.
2. The order of the differential equation is `2`.

Which of the above statements is/are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Given differential equation,

`(d^2y)/(dx^2) + cos (( dy)/(dx)) = 0`

I. Since, the given differential equation is not a polynomial

equation in its derivatives, then its degree is not defined.

II. Since, the highest order derivative in the given polynomial

is `2`.

Hence, the order of the differential equation is `2`.
Correct Answer is `=>` (C) Both 1 and 2
Q 1701612528

What is the degree of the differential equation

`((d^3y)/(dx^3))^(3//2) = ((d^2y)/(dx^2))^(2) ?`
NDA Paper 1 2014
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Consider the given differential equation,

`((d^3y)/(dx^3))^(3//2) = ((d^2y)/(dx^2))^(2)` ............(1)

We know that, in order to find degree, differential equation

should be free from fractional indices, therefore on

squaring both sides we get

`((d^3y)/(dx^3))^(3) = ((d^2y)/(dx^2))^(4)`

Since, the power of highest order derivative is `3`, therefore

degree `= 3`.
Correct Answer is `=>` (C) `3`
Q 1721812721

What is the solution of the equation `ln((dy)/(dx)) + x = 0`?

where, `C` is an arbitrary constant
NDA Paper 1 2014
(A)

`y + e^x = C`

(B)

`y - e^(-x) = C`

(C)

`y + e^(-x) = C`

(D)

`y - e^(x) = C`

Solution:

Consider the given differential equation

`ln ((dy)/(dx)) + x = 0`

`=> ln ((dy)/(dx)) = -x => (dy)/(dx) = e^(-x)`

On separating the variables, we get

` dy = e^(-x) dx`

On integrating both sides, we get

` int dy = int e^(-x) dx`

` => y = (e^(-x))/(-1) + C => y = -e^(-x) + C`

`=> y + e^(-x) = C`
Correct Answer is `=>` (C) `y + e^(-x) = C`
Q 1771012826

Eliminating the arbitrary constants `B` and `C` in the
expression `y = 2/(3C) (Cx -1)^( 3//2) + B,` we get
NDA Paper 1 2014
(A)

` x [ 1 + ( (dy) /(dx))^2 ] = (d^2y)/(dx^2)`

(B)

`2x ((dy) /(dx)) (d^2y)/(dx^2) = 1+ ((dy) /(dx)) ^2`

(C)

`((dy) /(dx)) (d^2y)/(dx^2) = 1`

(D)

` ( (dy) /(dx))^2 + 1 = (d^2y)/(dx^2)`

Solution:

Consider the given expression is

` y = 2 /(3c) (Cx - 1)^(3//2) + B`

On differentiating both sides w.r.t. x, we get

`(dy) /(dx) = 2 /(3c) . 3/2 (Cx - 1)^(1//2) . C + 0 = (Cx - 1)^(1//2)`

On squaring both sides, we get

`( (dy) /(dx))^2 = Cx - 1`

` => ( (dy) /(dx))^2 + 1 = Cx` ..............(1)

Now, on differentiating w.r.t. x, we get

` 2 (dy) /(dx) . (d^2y)/(dx^2) =C`

From Eq. (1),

`( (dy) /(dx))^2 + 1 = 2x (dy) /(dx) (d^2y)/(dx^2)`
Correct Answer is `=>` (B) `2x ((dy) /(dx)) (d^2y)/(dx^2) = 1+ ((dy) /(dx)) ^2`
Q 2329623511

What is the order of the differential equation `((dy)/(dx))^2+(dy)/(dx) - sin^2y = 0?`
NDA Paper 1 2013
(A)

`1`

(B)

`2`

(C)

`3`

(D)

Undefined

Solution:

Given differential equation is

`((dy)/(dx))^2+(dy)/(dx) - sin^2y = 0`

The highest order derivative, present in the differential equation is


`((dy)/(dx))` Therefore, its order is one
Correct Answer is `=>` (A) `1`
Q 2309623518

`y= 2cosx + 3sinx` satisfies which of the following
differential equations?

`I . (d^2y)/(dx^2) +y = 0`

`II . ((dy)/(dx))^2+(dy)/(dx) = 0`

Select the correct answer using the codes given below.
NDA Paper 1 2013
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Given differential equation is
`y = 2cosx + 3sinx` ..............(i)

Now `(dy)/(dx) = -2sinx+3cosx`

Again ` (d^2y)/(dx^2) = -2cosx-3sinx`


` = -2(2cosx+3sinx)`

` = -y` [fromEq.(i)]

`therefore (d^2y)/(dx^2)+y = 0`

So, only Statement I is correct.
Correct Answer is `=>` (A) Only I
Q 2359723614

The differential equation of all circles whose
centres are at the origin is
NDA Paper 1 2013
(A)

`(dy)/(dx) = y/x`

(B)

`(dy)/(dx) = x/y`

(C)

`(dy)/(dx) = -x/y`

(D)

None of these

Solution:

The equation of all circles whose centres are atl the
origin is,

`(x -0)^2+(y-0)^ = r^2`

` => x^2+y^2 = r^2` ........(i)

On differentiating w.r.t. x, we get

`2x+2y (dy)/(dx) = 0`

`=> (dy)/(dx) = -x/y`

which is the required differential equation of all circles whose
centres are at the origin.
Correct Answer is `=>` (C) `(dy)/(dx) = -x/y`
Q 2329823711

What is the degree of the differential equation

`((d^4y)/(dx^4))^(3/5)-5(d^3y)/(dx^3)+6(d^2y)/(dx^2)-8(dy)/(dx)+5 = 0?`
NDA Paper 1 2013
(A)

`5`

(B)

`4`

(C)

`3`

(D)

`2`

Solution:

Given differential equation is

`((d^4y)/(dx^4))^(3/5)-5(d^3y)/(dx^3)+6(d^2y)/(dx^2)-8(dy)/(dx)+5 = 0`

Since, the highest exponent of the highest derivative is called
degree of a differential equation provided exponent of clach
derivative and the unknown variable appearing in the differential
equation is an non-negative integer


`therefore ((d^4y)/(dx^4))^(3/5) = 5(d^3y)/(dx^3)-6(d^2y)/(dx^2)+8(dy)/(dx)-5`


`=> ((d^4y)/(dx^4))^(3) ={ 5(d^3y)/(dx^3)-6(d^2y)/(dx^2)+8(dy)/(dx)-5}^5`

`therefore ` Required degree `= 3`
Correct Answer is `=>` (C) `3`
Q 2369823715

The general solution of the differential equation

`x(dy)/(dx)+y = 0` is
NDA Paper 1 2013
(A)

`xy = C`

(B)

`x = Cy`

(C)

`x + y = C`

(D)

`x^2 + y^2 = C`

Solution:

Given differential equation is


`x(dy)/(dx)+y = 0 => x(dy)/(dx) = -y`


`=> -(dy)/y = (dx)/x => int(dx)/x+int(dy)/y = 0`


On integrating both sides, we get

`logx+ logy= logC`

`=> log(xy) = logC - xy= C`
Correct Answer is `=>` (A) `xy = C`
Q 2319023810

The general solution of the differential equation

`log((dy)/(dx))+x = 0` is
NDA Paper 1 2013
(A)

`y=e^(-x) + C`

(B)

`y=-e^(-x) + C`

(C)

`y = e^(x) + C`

(D)

`y = -e^(x) + C`

Solution:

Given differential equation is


`log((dy)/(dx))+x = 0 => log((dy)/(dx)) = -x`

`=> (dy)/(dx) = e^(-x) => intdy = inte^(-x) * dx`

On integrating both sides, we get
`y = -e^(-x) +C`
which is the required general solution.
Correct Answer is `=>` (B) `y=-e^(-x) + C`
Q 2339023812

The differential equation of the curve `y = sinx` is

NDA Paper 1 2013
(A)

`(d^2y)/(dx^2)+y(dy)/(dx)+x = 0`

(B)

`(d^2y)/(dx^2)+y = 0`

(C)

`(d^2y)/(dx^2)-y = 0`

(D)

`(d^2y)/(dx^2)+x = 0`

Solution:

Given curve is
`y = sinx` ......(i)
On differentiating w.r.t. x, we get

`(dy)/(dx) = cosx` .......(ii)

Again, differentiating w.r.t. x, we get


`(d^2y)/(dx^2) = -sinx = -y` fromeq(i)

`=> y+(d^2y)/(dx^2) = 0`

`=> (d^2y)/(dx^2)+y = 0`

which is the required differential equation
Correct Answer is `=>` (B) `(d^2y)/(dx^2)+y = 0`
Q 2369123915

The degree and order respectively of the of the `(dy)/(dx) = 1/(x+y+1)` are
NDA Paper 1 2013
(A)

`1, 1`

(B)

`1,2`

(C)

`2, 1`

(D)

`2, 1`

Solution:

Given differential equation is `(dy)/(dx) = 1/(x+y+1) => (x+y+1) (dy)/(dx) = 1`

`=> y(dy)/(dx)+(1+x)(dy)/(dx) = 1`

Since, the order of highest differential coefficient (highest
derivative) appearing in a differential equation is the order of
differential equation.
The highest exponent of the highest derivative is called degree of
a differential equation provided exponent of each derivative and
the unknown variable appearing in the differential equation is a
non-negative integer.
. . Order= 1 and degree = 1
Correct Answer is `=>` (A) `1, 1`
Q 2319123919

What is the degree of the differential equation

`(d^3y)/(dx^3)+2((d^2y)/(dx^2))^2-(dy)/(dx)+y = 0?`
NDA Paper 1 2012
(A)

`6`

(B)

`3`

(C)

`2`

(D)

`1`

Solution:

The power of highest derivative is 1.
So, degree = 1
Correct Answer is `=>` (D) `1`
Q 2329134011

Consider a differential equation of order m and
degree n. Which one of the following pairs is not
feasible?
NDA Paper 1 2012
(A)

`(3, 2)`

(B)

`(2, 3/2)`

(C)

`(2, 4)`

(D)

`(2,2)`

Solution:

The pairs `( 2,3/2)` is not feasible, because the degree of
any differential equation cannot be rational type.
If so, then we use rationalisation and convert it into integer.
Correct Answer is `=>` (B) `(2, 3/2)`
Q 2309134018

The differential equation representing the family
of curves `y = a sin (lamdax + alpha)` is
NDA Paper 1 2012
(A)

`(d^2y)/(dx^2)+lamda^2y = 0`

(B)

`(d^2y)/(dx^2)-lamda^2y = 0`

(C)

`(d^2y)/(dx^2)+lamda^2y = 0`

(D)

None of these

Solution:

Given `y = a sin (lamdax + alpha)` ......(i)

On differentiating it w.r.t. x. we get


`(dy)/(dx) = d/(dx)( a sin (lamdax + alpha))`


`a lamda cos (lamdax+alpha)lamda`

` = alamdacos(lamda+alpha)`


Again, differentiating it w.r.t x, we get


`(d^2y)/(dx^2) = a lamda d/(dx) cos(lamdax+alpha)`

` = a lamda[-sin(lamdax+alpha)lamda`

` = -alamda^2 sin(lamdax+alpha)`

`=> (d^2y)/(dx^2) = -lamda^2y` [from eq(i)]

`(d^2y)/(dx^2)+lamda^2y = 0`
Correct Answer is `=>` (A) `(d^2y)/(dx^2)+lamda^2y = 0`
Q 2349234113

The differential equation `y (dy)/(dx)+x = a` where `a` is any constant, represents
NDA Paper 1 2012
(A)

a set of straight lines

(B)

a set of ellipses

(C)

a set of circles

(D)

None of these

Solution:

`y (dy)/(dx)+x = aydy+xdx = adx`

On integrating both sides, we get


`intydy+intxdx = intadx`


`=> y^2/2+x^2/2 = ax`

`=> x^2+y^2-2ax = 0`

which represents a set of circles
Correct Answer is `=>` (C) a set of circles
Q 2339434312

For the differential equation

`((dy)/(dx))^2-x(dy)/(dx)+y = 0`

which one of the following is not its solution?

NDA Paper 1 2012
(A)

`y =x- 1`

(B)

`4y =x^2`

(C)

`y = x`

(D)

`y =-x - 1`

Solution:

The given differential equation is
`((dy)/(dx))^2-x(dy)/(dx)+y = 0` ..........(i)



(a) `y = x - 1 => (dy)/(dx) = 1`

From Eq. (i), `(1)^2 - x (1) + (x - 1) = 1- x + x - 1 = 0`
So, `y = x -1` is a solution of Eq. (i).

(b) `4y = x^2 => y = x^2/4 => (dy)/(dx) = x/2`

From eq(i) `(x/2)^2-x(x/2)+(x^2/4)`


` = x^2/4-x^2/2+x^2/4 = x^2/2-x^2/2 = 0`

So `4y = x^2 ` is a solutions of eq(i)

(c) `y = x => (dy)/(dx) = 1`

From Eq(i) , `(-1)^2-x(1)+x = 1 ne 0`
So,` y =x` is not a solution of Eq. (i).

(d) `y = -x-1 => (dy)/(dx) = -1`

From Eq(i) , `(-1)^2-x(-1)+(-x-1)`

` = 1+x-x-1 = 0`

So `y = -x-1` is a solution of Eq(i)
Correct Answer is `=>` (C) `y = x`
Q 2309534418

What is the general solution of the differential
equation `x^2 dy + y^2 dx = 0?`

1976
(A)

`x + y = C`

(B)

`xy = C`

(C)

`C (x + y) = xy`

(D)

None of these

Solution:

`x^2 dy + y^2 dx = 0 => (dy)/y^2+(dx)/x^2 = 0`

On integrating, we get

`inty^(-2)dy+intx^(-2)dx = 0`


`=> y^(-2+1)/(-2+1)+x^(-2+1)/(-2+1) = -C_1`


`=> y^(-1)/(-1)+x^(-1)/(-1) = C_1`

`=> (-1)/y -1/x = C_1 => 1/x+1/y = C_1x+y = C_1xy`


` => 1/C_1(x+y) = xy`

`=> C(x+y) = xy ` Where `1/C_1 = C`
Correct Answer is `=>` (C) `C (x + y) = xy`
Q 2319734610

What is the general solution of the differential
equation `e^x tany dx + (1- e^x') sec^2 y dy = 0?`
NDA Paper 1 2012
(A)

`sin y = C (1 -e^x)`

(B)

`cos y = C(1-e^x)`

(C)

`cot y = C (1 -e^x)`

(D)

None of these

Solution:

Given `e^xtany dx+(1-e^x) sec^2ydy = 0`


`=> e^x/(1-e^x) * dx +(sec^2y)/(tany) *dy = 0`


On integrating, we get


`=> int(e^xdx)/(1-e^x)+int(sec^2y)/(tany)dy = 0`


`=> -log(1-e^x)+logtany = logC`

`=> logtany = logC+log(1-e^x) = logC(1-e^x)`


`=> tany = C(1-e^x)`
Correct Answer is `=>` (D) None of these
Q 2389834717

Which one of the following differential equations
is not linear?'
NDA Paper 1 2012
(A)

`(d^2y)/(dx^2)+4y = 0`

(B)

`x(dy)/(dx)+y = x^3`

(C)

`(x-y)^2(dy)/(dx) = 9`

(D)

`cos^2x (dy)/(dy)+y = tanx`

Solution:

A differential equation which is of the form


`(dy)/(dx)+py = Q` and `(d^2y)/(dx^2)+p(dy)/(dx)+Qy = R` is called a linear equation

(a) `(d^2y)/(dx^2)+4y = 0` is a linear equation.

(b) `x(dy)/(dx)+y = x^3 => (dy)/(dx)+y/x = x^2` is a linear equation

(d) `cos^2x (dy)/(dy)+y = tanx`

So `(dy)/(dx)+sec^2x * y = tan x * sec^2 x` is also a linear equation

While (c) `(x-y)(dy)/(dx) = 9` is not a linear equation
Correct Answer is `=>` (C) `(x-y)^2(dy)/(dx) = 9`
Q 2309034818

What is the solution of the differential equation'?

`(dy)/(
dx)+y/x = 0?`

NDA Paper 1 2012
(A)

`xy = C`

(B)

`x = Cy`

(C)

`y =Cx`

(D)

None of these

Solution:

Given, differential equation is `(dy)/(dx)+y/x = 0` compare with `(dy)/(dx)+Py = Q`

Here `P = 1/x` and `Q = 0`

Now `IF = e^(intpdx) = e^(int(dx)/x) = e^(logx) = x`


`y * (IF) = int(Q * (IF)dx+C)`

So, the required solution is,

`y*x = int0dx +C => xy = C`
Correct Answer is `=>` (A) `xy = C`
Q 2319134919

What is the degree of the differential equation

`y = x(dy)/(dx)+((dy)/(dx))^
(-1) ?`

NDA Paper 1 2012
(A)

`1`

(B)

`2`

(C)

`-1`

(D)

Does not exist

Solution:

Given, differential equation is

`y = x(dy)/(dx)+((dy)/(dx))^(-1) => y = x (dy)/(dx)+1/(dy//dx)`


`=> y((dy)/(dx)) = x((dy)/(dx))^2+1`


Required degree = Power of highest derivative = 2
Correct Answer is `=>` (B) `2`
Q 2309145018

What is the equation of the curve passing through
the point `( 0, pi/3)` satisfying the differential
equation `sin x cos y dx +cos x sin y dy = 0?`
NDA Paper 1 2011
(A)

`cosx cosy = sqrt3/2`

(B)

`sinx sin y = sqrt3/2`

(C)

`sinx sin y = 1/2`

(D)

`cosx cos y = 1/2`

Solution:

Given differential equation is
`sin x ·cos y dx + cos x ·sin y dy = 0`
`=> sin x · cos y dx = - cos x · sin y dy`
`=> tan x dx = - tan y dy`

`=> tan x dx + tan y dy = 0`

`=> inttanxdx +int tany dy = 0`


`=> log secx+log sec y = logC^(-1)`


`=> log(secx * sec y) = log(1/C)`

`=> 1/(cosx * cosy) = 1/C`

`=> cos x * cos y = C` ...........(i)

Since, it passes through the point `( 0 , pi/3)` then

`cos0 * cos (pi/3) = C`

`=> 1 * 1/2 = C => C = 1/2`

From Eq(i)

`cosx * cos y = 1/2`
Correct Answer is `=>` (D) `cosx cos y = 1/2`
Q 2369345215

What is the degree of the differential equation

`((d^3y)/(dx^3))^(2/3)+4-3((d^2y)/(dx^2))+5(dy)/(dx) = 0?`


NDA Paper 1 2011
(A)

`3`

(B)

`2`

(C)

`2/3`

(D)

Not defined

Solution:

Given differential equation



`((d^3y)/(dx^3))^(2/3)+4-3((d^2y)/(dx^2))+5(dy)/(dx) = 0`


`=> ((d^3y)/(dx^3))^(2/3) = 3((d^2y)/(dx^2))-5(dy)/(dx) -4`



`=> ((d^3y)/(dx^3))^2 = { 3((d^2y)/(dx^2))-5(dy)/(dx) -4}^3`



Required degree = Powers of the highest order = 2
Correct Answer is `=>` (B) `2`
Q 2339445312

What is the solution of the differential equation

`(dy)/(dx)+ sqrt((1-y^2)/(1-x^2)) = 0?`
NDA Paper 1 2011
(A)

`sin^(-1)y+sin^(-1)x = C`

(B)

`sin^(-1)y-sin^(-1)x = C`

(C)

`2sin^(-1)y+sin^(-1)x = C`

(D)

`2sin^(-1)y-sin^(-1)x = C`

Solution:

The given differential equation is

`(dy)/(dx)+ sqrt((1-y^2)/(1-x^2)) = 0`


`=> int1/sqrt(1-y^2)dy+int1/sqrt(1-x^2)dx = 0`


`=> sin^(-1)y+sin^(-1)x = C`
Correct Answer is `=>` (A) `sin^(-1)y+sin^(-1)x = C`
Q 2319545410

What is the differential equation of all parabolas
whose axes are parallel to `Y`.axis?
NDA Paper 1 2011
(A)

`(d^3y)/(dx^3) = 0`

(B)

`(d^2x)/(dy^2) = C`

(C)

`(d^3x)/(dy^3) = 1`

(D)

`(d^3y)/(dx^3) = C`

Solution:

The general equation of all parabolas whose axes are
parallel toY-axis, is
`y = Ax^2 + Bx + C = 0` ... (i)
where, `A, B` and `C` are arbitrary constants.
On differentiating Eq. (i) w.r.t. x, we get

`(dy)/(dx) = 2Ax+B` ...........(ii)

On differentiating Eq. (ii) w.r.t. x, we get


`(d^2y)/(dx^2) = 2A` ............(iii)


On differentiating Eq. (iii) w.r.t. x, we get

`(d^3y)/(dx^3) = 0`
Correct Answer is `=>` (A) `(d^3y)/(dx^3) = 0`
Q 2319645510

If the solution of the differential equation

`(dy)/(dx) = (ax+3)/(2y+f)` represents a circle, then what is the value of a?
NDA Paper 1 2011
(A)

`2`

(B)

`1`

(C)

`-2`

(D)

`-1`

Solution:

`(dy)/(dx) = (ax+3)/(2y+f)` (given)


`=> int(2y+f)dy = int(ax+3)dx`

`=> c+y^2+fy = (ax^2)/2+3x`


`=> -a/2x^2+y^2-3x+fy+C = 0`


This equation represents a circle, if
Coefficient of `x^2` = Coefficient of `y^2`


` => -1 = a/2 => a = -2`
Correct Answer is `=>` (C) `-2`
Q 2319745610

What does the differential equation `y dy + x = a`
represent?
NDA Paper 1 2011
(A)

A set of circles having .centre on the Y-axis

(B)

A set of circles hav•ng centre on the X-axis

(C)

A set of ellipses

(D)

A pair of straight lines

Solution:

Given equation is `y dy + x = a => intydy+intxdx = aintdx`



`=> y^2/2+x^2/2-ax = c/2 => x^2+y^2 -2ac = c`


This represents a circle whose centre is on the X-axis
Correct Answer is `=>` (B) A set of circles hav•ng centre on the X-axis
Q 2379745616

What is the degree of the differential equation


`(1+(dy)/(dx))^4 = ((d^2y)/(dx^2))^2 ?`

NDA Paper 1 2010
(A)

`1`

(B)

`2`

(C)

`4`

(D)

`8`

Solution:

The given differential equation is

`(1+(dy)/(dx))^4 = ((d^2y)/(dx^2))^2 `


From above it is clear that, the degree of given differential
equation is 2.
(`because` degree of differential equation = degree of the higher order
derivative)
Correct Answer is `=>` (B) `2`
Q 2309845718

What is the general solution of


`(1+e^x)ydy = e^xdx ?`
NDA Paper 1 2010
(A)

`y^2 = In [C^2(e^x + 1)^2 ]`

(B)

`dy= In [C(e^x + 1)]`

(C)

`y^2 = In [C(e^x + 1)]`

(D)

None of the above

Solution:

The given differential equation is



`(1+e^x)ydy = e^xdx `

`=> intydy = int((e^x)/(1+e^x))dx`


`=> y^2/2 = log(1+e^x)+logC`


`=> y^2 = 2log [ C(1+e^x)]`


`=> y^2 = log[C^2(1+e^x)^2]`
Correct Answer is `=>` (A) `y^2 = In [C^2(e^x + 1)^2 ]`
Q 2359045814

Which one of the following is the differential
equation to family of circles having centre .at the
origin?
NDA Paper 1 2010
(A)

`(x^2-y^2)(dy)/(dx) = 2xy`

(B)

`(x^2+y^2)(dy)/(dx) = 2xy`

(C)

`(dy)/(dx) = (x^2+y^2)`

(D)

`xdx+ydy = 0`

Solution:

The equation of family of circles having centres at the
origin is

`x^2+y^2 = r^2`


On differentiating w.r.t. x, we get
`2x dx + 2 y dy = 0`

`=> x dx + ydy= 0`
which is the required differential equation
Correct Answer is `=>` (D) `xdx+ydy = 0`
Q 2339145912

What does the solution of the differential
equation `x (dy)/(dx) = y` represent?
NDA Paper 1 2010
(A)

Family of straight lines through the origin

(B)

Family of circles with their centres at the origin

(C)

Family of parabolas with their vertices at the origin

(D)

Family of straight lines having slope 1 and not passing through the origin

Solution:

`because x(dy)/(dx) = y => int1/ydy = int1/xdx`



`=> logy = logx+logC => y = xC`


which is a family of straight lines passing through the origin.
Correct Answer is `=>` (A) Family of straight lines through the origin
Q 2339256112

What is the differential equation to family of
parabolas having their ,vertices at the origin and
foci on the X-axis?
NDA Paper 1 2010
(A)

`y = 2-xy'`

(B)

`X = 2yy'`

(C)

`xy = y'`

(D)

`X = yy'`

Solution:

Required equation of parabola is
`y^2 = 4ax`......(i)
On differeniiating w.r.t. x, we get

`2yy' = 4a => 1/2yy' = a`


On putting this value of a in Eq. (i), we get


`y^2 = 4/2 yy'x => y = 2xy'`
Correct Answer is `=>` (A) `y = 2-xy'`
Q 2389256117

What is the solution of the differential equation

`3e^xtanydx+(1+e^x)sec^2ydy = 0`
NDA Paper 1 2010
(A)

`(1 + e^x)tan y = c`

(B)

`(1 + e^x)^3 tan y = c`

(C)

`(1 + e^x)^2 tan y = c`

(D)

`(1 + e^x) sec^2 y = c`

Solution:

`3e^xtanydx+(1+e^x)sec^2ydy = 0`


`=> int(3e^x)/(1+e^x)dx+int(sec^2y)/(tany)dy = 0`


`=> 3log(1+e^x)+logtany = logC`


`=> log(1+e^x)^3 tany = logC`

`=> (1+e^x)^3 tany = C`
Correct Answer is `=>` (B) `(1 + e^x)^3 tan y = c`
Q 2329356211

What is the differential equation for `y^2 = 4a(x-a) ?`


NDA Paper 1 2010
(A)

`yy'- 2xyy' + y^2 = 0`

(B)

`yy'(yy'+2x)+y^2 = 0`

(C)

`yy'(yy'-2x)+y^2 = 0`

(D)

`yy'-2xyy'+y = 0`

Solution:

`because y^2 = 4a(x-a)` ..........(i)

On differentiating w.r.t. x, we get

`y^2 = 4((yy')/2)(x-(yy')/2)`

`=> y^2 = yy'(2x-yy') => yy'(yy'-2x)+y^2 = 0`
Correct Answer is `=>` (C) `yy'(yy'-2x)+y^2 = 0`
Q 2318480300

What does an equation of the first degree
contammg one arbitrary parameter passing
through a fixed point represent?
NDA Paper 1 2009
(A)

Cirle

(B)

Straight line

(C)

Parabola

(D)

Ellipse

Solution:

The required equation represents a straight line.
(by definition).
Correct Answer is `=>` (B) Straight line
Q 2319356219

What is the solution of the differential equation

`-cosec^2(x+y)dy = dx ?`


NDA Paper 1 2008
(A)

`y - C = sin (x + y)`

(B)

`x - C =sin (x + y)`

(C)

`y- C = tan (x + y)`

(D)

None of these

Solution:

`-cosec^2(x+y)dy = dx `


`(dy)/(dx) = -sin^2(x+y)`

Put `x+y = t => 1+(dy)/(dx) = (dt)/(dx)`

`therefore (dt)/(dx)-1 = -sin^2(t)`

`=> (dt)/(dx) = 1-sin^2t`

`=> (dt)/(dx) = cos^2 t`


`=> intsec^2 tdt = intdx`

`=> tant = x-C`

`=> tan(x+y) = x-C`
Correct Answer is `=>` (D) None of these
Q 2379456316

What are the order and degree respectively of the
differential equation


`{((d^4y)/(dx^4))^3}^(2/3) -7x ((d^3y)/(dx^3))^2 = 8?`
NDA Paper 1 2008
(A)

`3 ,2`

(B)

`4, 3`

(C)

`4 ,2`

(D)

`3 ,3`

Solution:

`{((d^4y)/(dx^4))^3}^(2/3) -7x ((d^3y)/(dx^3))^2 = 8`


`=> ((d^4y)/(dx^4))^2-7x((d^3y)/(dx^3))^2 = 8`


So, the order and degree respectively of the differential equation
are 4 and 2.
Correct Answer is `=>` (C) `4 ,2`
Q 2349556413

What is the solution of the differential equation

`xdy -ydx = xy^2dx`
NDA Paper 1 2008
(A)

`yx^2+2x = 2Cy`

(B)

`y^2x+2y = 2Cx`

(C)

`y^2x^2+2x = 2Cy`

(D)

None of these

Solution:

`because xdy-ydx = xy^2dx`


`=> (xdy-ydx)/y^2 = xdx`


`=> intd(x/y) = -intxdx`

`=> x/y = -x^2/2+C`

`=> 2x+x^2y = 2Cy`
Correct Answer is `=>` (A) `yx^2+2x = 2Cy`
Q 2309556418

What does the solution of the differential
equation `x dy - y dx = 0` represent?
NDA Paper 1 2008
(A)

Rectangular hyperbola

(B)

Strai!Jlt line passing through origin

(C)

Parabola whose vertex is at origin

(D)

Circle whose centre is at origin

Solution:

`because xdy-ydx = 0`


`=> xdy = ydx => int1/ydy = int1/xdx`


`=> logy = logx+logC => logy = logCx`


`=> y = Cx`

Thus, tile solution of the differential equation represents a
straight the passing through origin.
Correct Answer is `=>` (B) Strai!Jlt line passing through origin
Q 2339656512

What is the order of the differential equation

`(dy)/(dx)+y = 1/((dy)/(dx)) ?`
NDA Paper 1 2008
(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`(dy)/(dx)-y = 1/((dy)/(dx))`


`=> ((dy)/(dx))^2+y((dy)/(dx)) = 1`

Thus, the order of the differential equation is 1
Correct Answer is `=>` (C) `1`
Q 2389656517

Rate of growth of bacteria is proportional to the
number of bacteria present at that time. If x is the
number of bacteria present at any instant t, then
which one of the following is correct?
(take proportional constant equal to 1.)
NDA Paper 1 2008
(A)

`x =log t`

(B)

`x = Ce^t`

(C)

`e^x = t`

(D)

`x = sqrt t`

Solution:

Rate of growth of bacteria oc The number of bacteria
present at that time


`=> (dx)/(dt) prop x`


`=> (dx)/(dt) = xK` (Where K = constant)


`=> int1/xdx = K intdt => logx = tK+logC`


`log(x/C) = KCt => x = Ce^(kt)`

Given, proportional constant, `K = 1`


`therefore x = Ce^t`
Correct Answer is `=>` (B) `x = Ce^t`
Q 2319756610

What is the equation of the curve whose slope at
any point is equal to `2x` and which passes through
the origin?
NDA Paper 1 2008
(A)

`y (1-x) = x^2`

(B)

`y^2(1+x^2) = x^4`

(C)

`y^2 = (x+1)^2`

(D)

`y = x^2`

Solution:

Given. slope at point `(x, y) = 2x`


`therefore (dy)/(dx) = 2x => intdy = 2intxdx`


`=> y^2 = x^2+C`


Since, it is passes through the origin.
`0=0+C => C = 0`
So, the required equation of curve is `y = x^2`
.
Correct Answer is `=>` (D) `y = x^2`
Q 2369756615

What is the solution of the differential equation

`(dy)/(dx) = xy+x+y+1?`
NDA Paper 1 2008
(A)

`y = x^2/2+x+C`

(B)

`log(y+1) = x^2/2+x+C`

(C)

`y = x^2+x+C`

(D)

`log(y+1) = x^2+x+C`

Solution:

The given differential equation is

`(dy)/(dx) = xy+x+y+1`


`(dy)/(dx) = (x+1)(y+1)`


`=> int1/(1+y)dy = int(x+1)dx`


`log(1+y) = x^2/2+x+C`
Correct Answer is `=>` (B) `log(y+1) = x^2/2+x+C`
Q 2329856711

What are the order and the degree , respectively of the differential equation `((d^2y)/(dx^2))^(5/6) = ((dy)/(dx))^(1/3) ?`
NDA Paper 1 2008
(A)

`2 ,1`

(B)

`2 , 5`

(C)

`2 , 5/6`

(D)

`1 , 1/3`

Solution:

`((d^2y)/(dx^2))^(5/6) = ((dy)/(dx))^(1/3) => ((d^2y)/(dx^2))^5 = ((dy)/(dx))^(6/3)`



`((d^2y)/(dx^2))^5 = ((dy)/(dx))^2`


So, this is clear that the order and degree of given differential
equation are 2 and 5, respectively
Correct Answer is `=>` (B) `2 , 5`
Q 2389856717

A radioactive element disintegrates at a rate
proportional to the quantity of substance Q
present at any time t. What is the differential
equation of the disintegration?
NDA Paper 1 2007
(A)

`(dQ)/(dt) =-Q`

(B)

`(dQ)/(dt) = -KQ , K < 0`

(C)

`(dQ)/(dt) = -KQ , K > 0`

(D)

`(dQ)/(dt) = Q`

Solution:

Since, a radioactive element disintegrates at a rate
proportional to the quantity of substance Q present at any time t

`therefore (dQ)/(dt) prop -Q`


`=> (dQ)/(dt) = -KQ , K > 0`

which is the required differential equation.
Correct Answer is `=>` (C) `(dQ)/(dt) = -KQ , K > 0`
Q 2379056816

What is the solution of the differential equation

`(x+y)(dx-dy) = dx+dy ?`


NDA Paper 1 2007
(A)

`2 log (x + y) = C(y- x)`

(B)

`(y- x) + log (x + y) = C``

(C)

`(y/x)+[log(y/x)] = C`

(D)

None of these

Solution:

The given differential equation is
`(x + y)(dx - dy) = dx + dy`

`(x + y) dx - (x + y) dy = dx + dy`

`(x + y- 1) dx = (x + y + 1) dy`

`=> (dy)/(dx) = (x+y-1)/(x+y+1)`

Let `x+y = v` and `(dy)/(dx) = (dv)/(dx)-1`

`(dv)/(dx)-1 = (v-1)/(v+1) => (dv)/(dx) = (v-1)/(v+1)+1`


`(dv)/(dx) = (v-1+v+1)/(v+1) => (v+1)/(2v)dv = dx`

`=> 1/2int1dv+1/2int1/vdv = int1dx`

`=> 1/2v+1/2logv = x+C/2`

`x + y + log (x + y) = 2x + C`
`=> (y- x) + log (x + y) = C`
Correct Answer is `=>` (B) `(y- x) + log (x + y) = C``
Q 2369156915

What is the only solution of the initial value problem `y' = t (1 + y)` and `y(0) = 0?`
NDA Paper 1 2007
(A)

`y = -1+e^(t^2//2)`

(B)

`y = 1+e^(t^2//2)`

(C)

`y = -t`

(D)

`y = t`

Solution:

Given, equation is `y' = t (1 + y)`

`=> (dy)/(dt) = t(1+y)`

`=> int1/(1+y)dy = inttdt`


`=> log(1+y) = t^2/2+C`

`because y(0) = 0`

`therefore log1 = C => C = 0`


`therefore log(1+y) = t^2/2`

`=> y = -1+e^(t^2//2)`

whcih is the required solution
Correct Answer is `=>` (A) `y = -1+e^(t^2//2)`
Q 2349167013

'What is the differential equation of the curve

`y = ax^2+bx?`


NDA Paper 1 2007
(A)

`x^2(d^2y)/(dx^2)-2x(dy)/(dx)+2y = 0`

(B)

`x^2(d^2y)/(dx^2)-y((dy)/(dx))^2+2 = 0`

(C)

`(1-x^2)(d^2y)/(dx^2)-(y(dy)/(dx))^2 = 0`

(D)

None of these

Solution:

Given equation is

`y = ax^2+bx` ..........(i)

On differentiating, w.r.t. x, we get

`(dy)/(dx) = 2ax+b` ..........(ii)

On again differentiating w.r.t. x, we get

`(d^2y)/(dx^2) = 2a`


Put the value of a in Eq. (ii),


`(dy)/(dx)-x(d^2y)/(dx^2) = b`


Now put the value of a and b in Eq. (i).


`2y = x^2(d^2y)/(dx^2)+2x(dy)/(dx)-2x^2(d^2y)/(dx^2)`

` = -x^2(d^2y)/(dx^2)+2x(dy)/(dx)`

`x^2(d^2y)/(dx^2)-2x(dy)/(dx)+2y = 0`


which is the required solution
Correct Answer is `=>` (A) `x^2(d^2y)/(dx^2)-2x(dy)/(dx)+2y = 0`
Q 2309167018

What is the degree of the differential equation


`[1+((dy)/(dx))^2]^(3/2) = K(d^2y)/(dx^2) ?`


NDA Paper 1 2007
(A)

`4`

(B)

`3`

(C)

`2`

(D)

`1`

Solution:

The given differential equation is


`[1+((dy)/(dx))^2]^(3/2) = K(d^2y)/(dx^2) `


On squaring both the sides, we get



`[1+((dy)/(dx))^2]^(3) = K^2((d^2y)/(dx^2))^2 `


Thus, it is clear that degree of differential equation is 2.
Correct Answer is `=>` (C) `2`
Q 2359367214

What is the equation of the curve passing through
the origin and satisfying the differential equation
`4y = (y tan x +sec x) dx?`
NDA Paper 1 2007
(A)

`y = xcosx`

(B)

`ycosx = x`

(C)

`xy = cosx`

(D)

`ysinx = x`

Solution:

Given differential equation is
`dy=(ytanx +sec x)dx`

`=> (dy)/(dx) -ytanx = secx`

This is a linear equation of the form `(dy)/(dx)+Py = Q`


Therefore, `P = - tan x` and `Q = sec x`

`therefore` lnterat1ng factor(IF) ` = e^(intpdx) = e^(int(-tanxdx)) = e^(-(-logcosx))`

` = e^(logcosx) = cosx`

`therefore` The solution is y· {IF) `= intQ(IF)dx+C`

`ycosx = intcosx*secxdx+C`

`=> ycosx = int1dx+C`

`=> ycosx = x+C`

Since, this curve passes through (0, 0).

`therefore` `0=0+C => C=0`
So, the required equation of curve is
`ycos x = x`
Correct Answer is `=>` (B) `ycosx = x`
Q 2319567410

What is the solution of the differential equation

`(dy)/(dx) = sec(x+y)?`

NDA Paper 1 2007
(A)

`y + tan (x + y) = C`

(B)

`y-tan{(x+y)/2} = C`

(C)

`y+tan{(x+y)/2} = C`

(D)

`y +tan{(x-y)/2} =C`

Solution:

The given differential equation is


`(dy)/(dx) = sec(x+y)`

Let `x+y =u`

`1+(dy)/(dx) = (du)/(dx)`

`(dy)/(dx) = (du)/(dx)-1`

`therefore (du)/(dx)-1 = secu`

`=> (du)/(dx) = secu+1`

`=> int1/(1+secu)du = int1dx` `(because cosu = (1-tan^2u//2)/(1+tan^2u//2))`

`int(cosu)/(1+cosu)du = int1dx`

`int(1-tan^2u/2)du = int2dx`


`int(2-sec^2u/2)du = 2x+2C` `(because tan^2x = sec^2x-1)`



`=> 2u-2tanu/2 = 2x+2C`

`=> 2y+2x-2tan((x+y)/2) = 2x+2C`

`y -tan((x+y)/2) = C`
Correct Answer is `=>` (B) `y-tan{(x+y)/2} = C`
Q 2329567411

For what value of `k`, does the differential equation
`(dy)/(dx) = ky` represent the law of natural decay?
NDA Paper 1 2007
(A)

`-5`

(B)

`0`

(C)

`0.01`

(D)

`(10)^(-1)`

Solution:

`because (dy)/(dx) = ky`


`=> int(dy)/y = intkdx`


`=> logy = kx`

`=> k = (logy)/x`

(where, k = negative constant)

From above it is clear that the given differential equation
represents the value of natural decay for negative values of k.
Correct Answer is `=>` (A) `-5`
Q 2389567417

What is the degree of the differential equation

`K^2(d^2y)/(dx^2) = [1+((dy)/(dx))^3]^(3/2)?`

where, K is a constant
NDA Paper 1 2007
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

The given differential equation is `K^2(d^2y)/(dx^2) = [1+((dy)/(dx))^3]^(3/2)?`


`=> K^2((d^2y)/(dx^2))^2 = [1+((dy)/(dx))^3]^3`

From above it is clear that the degree of given differential
equation is 2
Correct Answer is `=>` (B) `2`
 
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