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Must Do Problems of Differential Equations for NDA

Must Do Problems of Differential Equations for NDA

Must Do Problems of Differential Equations for NDA
Q 2562080835

The degree and order of the differential equation

`y = x ((dy)/(dx))^2 +((dx)/(dy))^2` are respectively
WBJEE 2011
(A)

`1,1`

(B)

`2,1`

(C)

`4,1`

(D)

`1,4`

Solution:

Given differential equation is

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

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

Here, degree is `4` and order is `1`.
Correct Answer is `=>` (C) `4,1`
Q 2855178964

The family of curves `y = Ae^(3x) + Be^(5x)`, where `A` and `B` are parameters.
What is the differential equation of the given family of curves?

(A)

`(d^2y)/(dx^2) - 8 (dy)/(dx) + 15 y = 0`

(B)

`(d^2y)/(dx^2) + 8 (dy)/(dx) + 15 y = 0`

(C)

`(d^2y)/(dx^2) + 8 (dy)/(dx) - 15 y = 0`

(D)

None of these

Solution:

Given, `y = Ae^(3x) + Be^(5x)` ... (i)

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

get ` (dy)/(dx) = 3 Ae^( 3 x) + 5 Be^(5x)`

`=> y_1 = 3Ae^(3x) + 5 Be^(5x)` ... (ii)

` [ ∵ (dy)/(dx) = y_1 ]`

`=> y_2 = 9Ae^(3x) + 25 Be^(5x)` ... (iii)

Eliminating A and B in Eqs. (i), (ii) and

(iii), we get

`| ( e^(3x) , e^(5x) , -y ) , (3 e^(3x) , 5 e^(5x) , -y_1), ( 9 e^(3x) ,25 e^(5x) , -y_2 ) | = 0`

`=> (- l) e^(3x) · e^(5x) | (1,1,y),(3,5,y_1),(9 , 25 , y_2) | = 0`

` => | ( 1,1,y), ( 3 ,5, y_1) ,( 9 , 25 , y_2) | = 0`

`=> 1 (2y_2 - 16 y_1) + y ( 48 - 18) = 0`

`=> y_2 - 8 y_1 + 15 y = 0` ... (iv)
Correct Answer is `=>` (A) `(d^2y)/(dx^2) - 8 (dy)/(dx) + 15 y = 0`
Q 2456545474

If `y^ 2 = P(x)` be a cubic polynomial, then `2 d/(dx) (y^3 (d^2 y)/(dx^2))` is equal to
UPSEE 2010
(A)

` P(x)`

(B)

`P(x)p'(x)`

(C)

`P (x) P'''(x)`

(D)

constant

Solution:

`P' (x) = 2y y', P ' ' (x) = 2y y' ' + 2y'^2`

`=> P' ' ' (x) = 2y y ''' + 6y' y' '`

`:. 2d/(dx) (y^3y' ') = 2[y^3y' ' + 3y^2y' y' ']`

`= y ^2 [2y y ' ' + 6y' y"]`

`=P(x) P' ' ' (x)`
Correct Answer is `=>` (C) `P (x) P'''(x)`
Q 2553380244

The solution of the differential equation `sec^2 x tan y dx + sec^2 y tan x dy = 0` is :
UPSEE 2013
(A)

`tan y tan x = c`

(B)

`(tany)/(tanx) = C`

(C)

`(tan^2 x) /(tany) = C`

(D)

None of these

Solution:

Given differential equation is

`sec^2 x tan y dx + sec^2 y tan x dy = 0`

`=> (sec^2 x dx)/(tanx) = (-dy sec^2 y)/(tany)`


On integrating both sides


`int (sec^2 x dx)/(tanx) = - int ( sec^2 y)/(tany) dy`


`=> log (tan x) +log tan y =log c`


`=> tan x tan y = c`
Correct Answer is `=>` (A) `tan y tan x = c`
Q 2835467362

Solution of differential equation ` ( x + y (dy)/(dx) )/( y - x (dy)/(dx) ) = ( x cos^2 (x^2 + y^2))/y^3 ` is equal to

(A)

`tan (x^2 + y^2) = x^2/y^2 + C`

(B)

`cot (x^2 + y^2) = x^2/y^2 + C`

(C)

`tan (x^2 + y^2) = y^2/x^2 + C`

(D)

`cot (x^2 + y^2) = x^2/y^2 + C`

Solution:

The given equation can be written as

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

` = y^2 . x/y^3 cos^2 (x^2 + y^2)`

`=> sec^2 (x^2 + y^2) 1/2 d (x^2 + y^2)`

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

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

[on integrating]

` :. tan (x^2 + y^2) = x^2/y^2 + C`
Correct Answer is `=>` (A) `tan (x^2 + y^2) = x^2/y^2 + C`
Q 2502034838

If `sqrt y= cos^(-1) x`, then it satisfies the differential equation `(1-x^2) (d^2y)/(dx^2)-x(dy)/(dx)=c` where `c` is equal to
WBJEE 2014
(A)

`0`

(B)

`3`

(C)

`1`

(D)

`2`

Solution:

Given, `sqrt y =cos^(-1) x => y =(cos^(-1) x)^2`

On differentiat:mg both sides w.r.t `x`, we get

`(dy)/(dx)=2(cos^(-1) x) xx (-1)/(sqrt(1-x^2))`

Again, differentiat.ing both sides w.r.t. `x`, we get

`(d^2x)/(dx^2) =-2 [(sqrt(1-x^2) xx (-1)/(sqrt(1-x^2))-cos^(-1) x xx (1/2) ((-2x))/((1-x^2)^(1//2)))/((sqrt(1-x^2))^2)]`

`(d^2y)/(dx^2) =[(2-(2xcos^(-1)x)/((1-x^2)^(1//2)))/((1-x^2))]`

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

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

But it is given

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

`:. c=2`
Correct Answer is `=>` (D) `2`
Q 2552034834

The integrating factor of the differential equation `(1+x^2) (dy)/(dx)+y=e^(tan^(-1) x)` is
WBJEE 2014
(A)

`tan^(-1) x`

(B)

`1+x^2`

(C)

`e^(tan^(-1) x)`

(D)

`log_e(1+x^2)`

Solution:

Given equation is `(1+x^2) (dy)/(dx)=e^(tan^(-1) x)`

or it can be rewritten as

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

It is a linear dlfferential equation of the form of

`(dy)/(dx)+Py=Q`

where, `P=1/(1+x^2)` and `Q=(e^(tan^(-1) x))/(1+x^2)`

`(dy)/(dx)+Py=Q`

where, `P = 1/(1+x^2)` and `Q=(e^(tan^(-1) x)/(1+x^2))`

`:.` Integrating factor `IF=e^(int Pdx)`

`=e^(int 1/(1+x^2))=e^(tan^(-1) x)`
Correct Answer is `=>` (C) `e^(tan^(-1) x)`
Q 2559078814

If `m` and `n` are the order and degree of the differential equation
` ( (d^2y)/(dx^2) )^5 + 4 ( (d^2y)/(dx^2) )^3/ (((d^3y)/(dx^3)) ) + (d^3 y)/(dx^3) = x^2 - 1` , then
BCECE Mains 2015
(A)

`m = 3, n = 3`

(B)

`m = 3, n = 2`

(C)

`m = 3, n = 5`

(D)

`m = 3, n = 1`

Solution:

Given differential equation can be rewritten as

` ( (d^3y)/(dx^3) )^2 - (x^2 - 1) (d^3 y)/(dx^3) + ((d^3 y)/(dx^3)) ((d^2 y)/(dx^2))^5`

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

Clearly, its order is `3` and degree is `2`.

`:. m = 3` and `n = 2`
Correct Answer is `=>` (B) `m = 3, n = 2`
Q 2449356213

The differential equation representing the family of curves `y^ 2 = 2c ( x + sqrt c )`, where `c` is a positive perimeter, is of
BCECE Stage 1 2011
(A)

order 1, degree 3

(B)

order 2, degree 2

(C)

degree 3, order 3

(D)

degree 4, order 4

Solution:

Given family of curves

`y^2=2c(x+sqrt c)`.............(i)

On differentiating both sides, we get

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

From Eq. (i),

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

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

On squaring both sides, we get

`(y^2-2xy (dy)/(dx))^2=4 (y (dy)/(dx))^3`

So, order 1 and degree 3.
Correct Answer is `=>` (A) order 1, degree 3
Q 1643801743

If `phi (x)` is a differentiable function, then the solution of
the differential equation `dy + {y phi' (x)- phi (x) phi' (x)}dx = 0`, is
BITSAT 2015
(A)

`y={phi(x) - 1} + Ce^(-phi(x))`

(B)

`y phi(x) = {phi(x)}^2+C`

(C)

`ye^(phi(x)) = phi(x) e^(phi(x)) +C`

(D)

`y-phi(x)e^(-phi(x))`

Solution:

We have, `dy + {y phi' (x)- phi (x) phi' (x)} dx = 0`

`=> (dy)/(dx) + phi' (x) y = phi (x) phi' (x)`

which is a linear differential equation with

`If = e^(int phi'(x)dx) = e^(phi(x))`

`:.` Solution is `y.e^(phi(x)) = int phi(x).phi' (x) e^(phi(x))dx+C`\

` => y.e^(phi(x)) = int phi underset I (x) . e^(phi(x)) phi' underset (II) (x) dx +C`

`=> y.e^(phi(x)) = phi (x) e^(phi(x)) - int phi' (x) e^(phi(x)) dx+C`

` => y.e^(phi(x)) = phi(x) e^(phi(x)) + C`

` => y = {phi(x) -1} + Ce^(-phi(x))`
Correct Answer is `=>` (A) `y={phi(x) - 1} + Ce^(-phi(x))`
Q 1570191016

The solution of the differential equation `xy^2 dy - (x^3 + y^3) dx = 0` is
BITSAT 2008
(A)

`y^3 = x^3 log cx`

(B)

`y^3 = 3x^3 log cx`

(C)

`x^3 = 3x^3 log cx`

(D)

`y^3 = 3x^3 log x`

Solution:

Given differential equation can be rewritten as

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

It is a homogeneous differential equation.

Put `y = vx`

`dy/dx = v+x (dv)/dx `

`v+x (dv)/dx = (x^3+v^3x^3)/(x^3v^2)`

`v+x (dv)/dx = (1+v^3)/v^2`

`x (dv)/dx = 1/v^2`

`v^2 dv= dx/x`

On integrating both sides, we get

`v^2/3= log x + log c`

`1/3 (y/x)^3= log x + log c`

`y^3 = 3x^3 log cx`
Correct Answer is `=>` (B) `y^3 = 3x^3 log cx`
Q 2509680518

The solution of the differential equation
` x = 1 + xy (dy)/(dx) + (xy)^2/(2!) ((dy)/(dx))^2 + (xy)^3/(3!) ((dy)/(dx))^3 + ....` is
BCECE Mains 2015
(A)

`y = log_e (x) + C`

(B)

`y = (log_e x)^2 + C`

(C)

` y = pm sqrt ( (log_e x)^2 + 2C )`

(D)

`xy = x^y + K`

Solution:

We have, ` x = e^( xy (dy)/(dx) )`

` = log x = xy (dy)/(dx) `

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

` => y dy = logx d ( log x)`

On integrating, we get

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

`=> y^2 = (log_e x)^2 + 2C`

` => y = pm sqrt ( (log_e x)^2 + 2C )`
Correct Answer is `=>` (C) ` y = pm sqrt ( (log_e x)^2 + 2C )`
Q 1571834726

The degree of the differential equation

`y(x) =1 + (dy)/(dx) + 1/(1 . 2) ((dy)/(dx))^2 + 1/(1 . 2 .3) ((dy)/(dx))^3 +.....` is
BITSAT 2005
(A)

`2`

(B)

`3`

(C)

`1`

(D)

None of these

Solution:

Given that,

`y(x) =1 + (dy)/(dx) + 1/(1 . 2) ((dy)/(dx))^2 + 1/(1 . 2 .3) ((dy)/(dx))^3 +.....`


`y(x) =1 + 1/(1 ! )(dy)/(dx) + 1/(2 !) ((dy)/(dx))^2 + 1/(3 ! ) ((dy)/(dx))^3 +.....`

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

Taking log on both sides, we get

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

`:.` The degree of this equation is 1.
Correct Answer is `=>` (C) `1`
Q 2431334222

The solution of the differential equation

`(dy)/(dx)+sin((y+x)/2)+sin((y-x)/2)=0 ` is
UPSEE 2015
(A)

`log tan(y/2)=C-2 sin x`

(B)

`log tan (y/4)=C-2 sin(x/2)`

(C)

`log tan (y/2+pi/4)=C-2 sin x`

(D)

`log tan (y/2+pi/4)=C-2 sin (x/2)`

Solution:

Given differential equation can be rewritten as

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

`=2 cos(x/2)sin(-y/2)`

`=-2 cos(x/2) sin(y/2)`

`=> int (dy)/(2 sin(y/2))=-int cos(x/2) dx`

`=> 1/2 int cosec (y/2)dy=-(sin(x/2))/((1/2))+C`

`=> 1/2 (log(cosec(y/2)-cot(y/2)))/((1/2))`

`=> - (sin (x/2))/(1/2)+C`

`=> log tan (y/4)`

`=-2 sin(x/2)+C`
Correct Answer is `=>` (B) `log tan (y/4)=C-2 sin(x/2)`
Q 2304712658

Solution of the differential equation
`(dy)/(dx) + ay = e^(mx)` is :
BITSAT Mock
(A)

`(a + m) y = e^(mx) + ce^(−ax)`

(B)

`y = e^(mx) + ce^(−ax)`

(C)

`(a + m) y = e^(mx) + c`

(D)

`ye^(ax) = me^(mx) + c`

Solution:

The equation `(dy)/(dx) + ay = e^(mx)` is of the form

` (dy)/(dx) + Py = Q` ...(1)

Here `P = a, Q = e^(mx)`

`∴ I.F. = e^(∫ Pdx)`

`= e^(∫ adx) = e^(ax)`

now solution is

`y × (I.F.) = ∫ {Q × (I.F.)} dx + c`

`⇒ ye^(ax) = ∫ (e^(mx) . e^(ax)) dx + c`

`⇒ ye^(ax) = ∫ e^((m + a) x) dx + c`

`⇒ ye^(ax) = e^((a + m) x)/(a + m) + c`

`⇒ (a + m) ye^(ax) = e^(ax) . e^(mx) + c`

`⇒ (a + m) y = e^(mx) + ce^(−ax)`

Which is the required solution.
Correct Answer is `=>` (A) `(a + m) y = e^(mx) + ce^(−ax)`
Q 2426512471

The solution of differential equation
`(x^2 + y^2) - 2xy (dy)/(dx) = 0` is
UPSEE 2014
(A)

`x^2 + y^2 = xC`

(B)

`x^2 - y^2 = xC`

(C)

`x^2 + y^2 = C`

(D)

`x^2 - y^2 = C`

Solution:

Given differential equation is

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

which is homogeneous.

(`∵` degree of each term is same i.e., `2`)

It can be rewritten as

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

Put `y = vx => (dy)/(dx) = v + x (dv)/(dx)`, so that the

differential equation becomes

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

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

` => int (2v)/(1 - v^2) dv = int 1/x dx`

`=> - log (1 - v^2) = log | x| - log |C |`

`=> x (1- v^2) = C`

`=> (x^2 - y^2)/x = C quad ( ∵ v = y/x)`

Hence, `x^2 - y^2 = xC` is the required solution.
Correct Answer is `=>` (B) `x^2 - y^2 = xC`
Q 2405223168

The solution of the differential equation

`x dy - y dx = (sqrt (x^2 + y^2) )dx` is
UPSEE 2012
(A)

`y - sqrt (x^2 + y^2) = Cx^2`

(B)

`y + sqrt (x^2 + y^2) = Cx^2`

(C)

`y + sqrt(x^2 + y^2) +Cx^2 = 0`

(D)

none of the above

Solution:

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

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

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

It is homogeneous equation.

Put `y = vx => (dy)/(dx) = v + x (dv)/(dx)`

`v+x (dv)/(dx) = (vx + x sqrt (1+v^2) )/x = (v + sqrt (1+v^2) )/1`

`=> x (dv)/(dx) = sqrt(1+v^2)`

`=> int (dv)/(sqrt (1+v^2) ) = int (dx)/x`

`=> log (v + sqrt (1+v^2) ) = log x + log C`

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

`=> log (y + sqrt (x^2 + y^2) ) - log x = log x + log C`


`=> log (y + sqrt (x^2 +y^2) ) = log (x^2 C)`

`=> y + sqrt (x^2+y^2) = x^2 C`
Correct Answer is `=>` (B) `y + sqrt (x^2 + y^2) = Cx^2`
Q 2845256163

The differential equation of curves `y = Ae^(3x) + Be^(5x)`, where the family of A and B are arbitrary constants, is

(A)

`(d^2y)/(dx^2) + 8 (dy)/(dx) + 15 y = 0`

(B)

`(d^2y)/(dx^2) - 8 (dy)/(dx) + 15 y = 0`

(C)

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

(D)

None of these

Solution:

` y = Ae^(3x) + Be^(5x)` ... (i)

`:. y_1 = 3 Ae^(3x) + 5Be^(5x)` ...... (ii)

`y_2 = 9 Ae^(3x) + 25 Be^(5x)` ...... (iii)

Eliminating A and B from the above

three equations, we get

` | ( e^(3x) , e^(5x) , -y),( 3 e^(3x) , 5 e^(5x) , -y_1 ) , ( 9e^(3x) , 25 e^(5x) , -y_2) | = 0`


`=> - e^(3x) . e^(5x) | ( 1,1,y),(3,5,y_1),( 9,25 , y_2)| = 0`

On expanding, we get

`30y - 16y_1 + 2y_2 = 0`

or ` (d^2y)/(dx^2) - 8 (dy)/(dx) + 15 y = 0`
Correct Answer is `=>` (B) `(d^2y)/(dx^2) - 8 (dy)/(dx) + 15 y = 0`
Q 2521580421

The integrating factor of the differential equation `3x log_e x (dy)/(dx) + y = 2 log_ex` is given by
WBJEE 2012
(A)

`(log_e x)^3`

(B)

`log_e (log_e x)`

(C)

`log_e x`

(D)

`(log_e x)^(1//3)`

Solution:

Given, `3x log_e x (dy)/(dx) + y = 2 log_ex`

Dividing both sides by `3x log_e x`, we get

` (dy)/(dx) + 1/(3x log_e x) y = ( 2log_e x)/(3x log_e x)`

` => (dy)/(dx) + 1/(3x log_e x) y = 2/( 3x)`

which is linear form ` (dy)/(dx) + Py = Q`, where `P` and

`Q` are function of `x` and the integrating factor is

given by the following formula `e ^(int Pdx)`.

`:.` IF `= e^(int 1/(3x log_e x)) dx`

Put `log_e x = t => 1/x dx = dt`

`= e^(1//3) int (dt)/t = e^(1//3) log t`

` = e^(log t^(1//3)) = t^( 1//3) = (log_e x )^(1//3)`
Correct Answer is `=>` (D) `(log_e x)^(1//3)`
Q 2408101908

The differential equation of all circles
touching the axis of y at origin and centre on
the x-axis is given by
UPSEE 2009
(A)

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

(B)

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

(C)

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

(D)

None of the above

Solution:

The equation of circle touching the

y-axis at `(0, 0)` and centre lies on x-axis is

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

Let the required equation of circle be

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

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

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

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

` => 2xy (dy)/(dx) + x^2 - y^2 = 0`
Correct Answer is `=>` (D) None of the above
Q 2487391287

Differential equation of those circles which
passes through origin and their centres lie on
y-axis will be
UPSEE 2009
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

Equation of circle whose centre is `(0, f)` and

radius `f`, is

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

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

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

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

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

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

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

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

Equation of circle passing through origin and

centre lies on x-axis is `x^2 + y^2 - 2gx = 0`
Correct Answer is `=>` (B) `(x^2 - y^2) (dy)/(dx) = 2xy`
 
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