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Tricks & Tips of Differential Equations for NDA

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Finding order and degree of differential equation

- The order of a differential equation is the order of the highest order derivative present in the equation. - The degree of a differential equation is the power of the highest order derivative in the equation.
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 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 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 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 1515323260

The order and degree of the differential equation `\sqrt { \frac { dy }{ dx } } -4\frac { dy }{ dx } -7x=0` are:
BITSAT 2011
(A)

`1` and `\frac { 1 }{ 2 }`

(B)

`2` and `1`

(C)

`1` and `1`

(D)

`1` and `2`

Solution:

Given, `\sqrt { \frac { dy }{ dx } } -4\frac { dy }{ dx } -7x=0`

On squaring , we get

`\frac { dy }{ dx } =16 ( \frac { dy }{ dx } ) ^{ 2 }+49 x ^{ 2 }+56x\frac { dy }{ dx }`

Here, order is `1` and degree is `2`.
Correct Answer is `=>` (D) `1` and `2`
Q 1581645527

The order and degree of the differential
equation `sqrt((dy)/(dx)) - 4 (dy)/(dx) - 7x = 0` are
BITSAT 2011
(A)

`1` and `1/2`

(B)

`2` and `1`

(C)

`1` and `1`

(D)

`1` and `2`

Solution:

Given, `sqrt((dy)/(dx)) - 4 (dy)/(dx) - 7x = 0`

On squaring, we get

`(dy)/(dx) = 16 ((dy)/(dx))^2 + 49x^2 + 56x (dy)/(dx)`

Here, order is `1` and degree is `2.`
Correct Answer is `=>` (D) `1` and `2`
Q 2845878763

The degree of the differential equation which satisfies ` sqrt(1 - x^2) + sqrt(1 - y^2) = a(x- y)` is
I. degree `= 1` II. degree `= 2` III. degree `= 3`

(A)

Only I

(B)

Only II

(C)

Only III

(D)

None of these

Solution:

We have,

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

Let `x = sin A` and `y = sin B`,

we get

`cos A + cos B = a (sin A - sin B)`

`A - B = 2 cot^(-1) a`

`sin^( -1) x - sin^(-1) y = 2 cot^(-1) a`

On differentiating w.r.t x, we get

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

Hence, it is a differential equation of `1` degree.
Correct Answer is `=>` (A) Only I
Q 2229680511

The order of the differential equation whose general solution is given by

`y = (C_1 + C_2) cos (x + C_3) + C_4 e^(x+ C_5)`

where `C_1 ,C_2 ,C_3 ,C_4 , C_5` are arbitrary constants, is
BITSAT Mock
(A)

`5`

(B)

`4`

(C)

`3`

(D)

`2`

Solution:

`y = A cos (x + B) + Ce^x` where `C_1 + C_2 = A` and `C_4 e^(C_5) = C`. Thus there are
only three arbitrary constants and hence the differential equation will
be of 3rd order.

`y' = -A sin (X+B) +Ce^x`

`y'' = -A cos (X+B) + Ce^x`

`y''' = A sin (X+B) +Ce^x`

`:. (y+y'')/(y'+y''') = (2 C e^x)/(2Ce^x) =1`

or `y''' - y'' + y' - y =0` is the differential equation.
Correct Answer is `=>` (C) `3`

Formation of differential equation : by eliminating one or two variable

- For any given differential equation, the solution is of the form `f(x,y,c_1,c_2,…….,c_n)=0` where x and y are the variables and `(c_1,c_2…….c_n)` are the arbitrary constants. To obtain the differential equation from this equation

we follow the following steps:-

Step 1: Differentiate the given function w.r.t to the independent variable present in the equation.

Step 2: Keep differentiating times in such a way that (n+1)equations are obtained.

Step 3: Using the (n+1) equations obtained, eliminate the constants `(c_1,c_2…….c_n) .`
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 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 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 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 2503245148

The equation of one of the curves whose slope at any point is equal to `y + 2x` is
WBJEE 2010
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

Given `(dy)/(dx) = y+2x`....(i)


Put `y +2x = z`

`=> (dy)/(dx)+2 = (dz)/(dx)`

`=> (dy)/(dx) = (dz)/(dx)-2` .......(ii)

From Eqs. (i) and (ii)

`(dz)/9dx)-2 = z`

`=> int(dz)/(z+2) = intdx`

`=> log(z+2) = x+c`

`=> log(y+2x) = x+c`


`=> y+2x+2 = e^(x+c)`

`=> y+2x+2 = e^x * e^c`

`=> y = 2[e^x-x-1]` Taking `e^c= 2`
Correct Answer is `=>` (B) `y= 2(e^x - x- 1)`
Q 2543512443

The curve `y =(cos x + y)^(1//2)` satisfies the differential equation
WBJEE 2014
(A)

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

(B)

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

(C)

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

(D)

None of these

Solution:

Given curve is

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

on differentiating both sides w.r.t. `x`, we get

`(dy)/(dx) =1/(2(cos x+y)^(1//2)) xx (-sin x+(dy)/(dx))`

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

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

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

again differentiating both sides, w.r.t `x` we get

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

`=> (2y-1) (d^2y)/(dx^2)+(dy)/(dx)(2 (dy)/(dx))+cos x=0`
Correct Answer is `=>` (A) `(2y-1)(d^2y)/(dx^2)+2 ((dy)/(dx))^2+cos x=0`
Q 2825356261

The differential equation of all parabolas having their axis of symmetry coinciding with the axis of `x`, is

(A)

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

(B)

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

(C)

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

(D)

None of these

Solution:

General equation of parabola having

X -axis as the axis of symmetry is

`y^2 = 4a(x - b)`

Now, differentiate and eliminate `a` and `b`.

On differentiating

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

Again , differentiate `((dy)/(dx))^2 + y . ( dy^2)/(dx^2) = 0`
Correct Answer is `=>` (A) ` y (d^2 y)/(dx^2) + ((dy)/(dx))^2 = 0`
Q 2574034856

The differential equation of all circles which passes through the origin and whose centre lies
on `y`-axis is
UPSEE 2008
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

The equation of all circles which passes through
the origin and whose centre lies on `y`-axis, is


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

On differentiating wrt `x`, we get

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

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

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

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

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

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

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

Which is required differential equation.
Correct Answer is `=>` (A) `(x^2 - y^2) (dy)/(dx) -2xy =0`
Q 1563591445

By eliminating the arbitrary constants `A` and `B` from `y=Ax^2+Bx`, we get the differential equation :
BITSAT 2012
(A)

`\ frac {d^3y}{dx^3}=0`

(B)

`x^2 \ frac {d^2y}{dx^2}-2x\ frac {dy}{dx}+2y=0`

(C)

`\ frac {d^2y}{dx^2}=0`

(D)

`x^2 \ frac {d^2y}{dx^2}+y=0`

Solution:

`y=Ax^2+Bx`.......(1)

`\Rightarrow y' =2Ax+B`.........(2)

and `y'' =2A\Rightarrow A=\ frac{y''}{2}`

Substitute `A` in (2) `\Rightarrow B={y'-y''x}`

Now substitute both A and B in (1)

`\Rightarrow 2y=y''x^2+2(y'-y''x)x=2y'x-y''x^2`

`\Rightarrow x^2\ frac{d^2y}{dx^2}-2x\ frac{dy}{dx}+2y=0`
Correct Answer is `=>` (B) `x^2 \ frac {d^2y}{dx^2}-2x\ frac {dy}{dx}+2y=0`

Finding solution of differential equation by differentiating the options

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`

Simplifying and Integrating

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 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`

Question using special relation

`xdy + y dx = d(xy)` `(xdy - ydx )/ x^2 = d(x/y)`
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 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 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`

solving Linear Differential equation

Solving Linear differential equations A differential equation should be form of `(dy)/(dx)+Py=Q` `IF= e^(int Pdx)` The solution of the given equation is given by `y*(IF)= int Q (IF) dx + C`
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 2815767660

The solution of the differential equation `(dy)/(dx) = y^2/(1 - 3xy)` is given by

(A)

`y^3 x = y^2/2 + C`

(B)

`y^3 = (xy^2)/2 + C`

(C)

` x =(1 + 2 Cy)/y^3`

(D)

` x = C/y^3`

Solution:

We have, ` (dy)/(dx) = y^2/(1 - 3xy)`

`=> (dy)/(dx) = (1 - 3xy)/y^2 => (dx)/(dy) = 1/y^2 - 3/y x`

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

The above is a linear differential

equation of the form

`(dx)/(dy) + P(y)x = Q(y)`

Here, ` P = 3/y ` and ` Q = 1/y^2`

`:. IF = e^( int 3/y dy) = e^(3 log x) = y^3`

The solution of Eq. (i) is given by

` xy^3 = int 1/y^2 . y^3 dy + C`

` :. xy^3 = y^2 /2 + C`
Correct Answer is `=>` (A) `y^3 x = y^2/2 + C`
Q 2488112007

The solution of the differential equation
`(e^(-2sqrtx) - y/sqrtx) (dx)/(dy) = 1` is given by
UPSEE 2009
(A)

`ye^(2sqrtx) = 2sqrtx + c`

(B)

`ye^(-2sqrtx) = sqrtx + c`

(C)

` y = sqrtx`

(D)

` y = 3 sqrtx`

Solution:

Given differential equation can be rewritten as

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

Here , ` P = 1/sqrtx , Q = e^(-2sqrtx)`

`:. IF = e^(int1/sqrtx dx) = e^(2sqrtx)`

Solution is

`ye^(2sqrtx) = int e^(2sqrtx) e^(-2sqrtx) dx`

`= int 1 dx => ye^(2sqrtx) = x + c`

Misc

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)`
 
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