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Differential Equations question

2019 · 12 Apr · Shift 2 · Q25
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  5. /2019 · 12 Apr · Shift 2 · Q25

Differential Equations question

2019 · 12 Apr · Shift 2 · Q25

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
The general solution of the differential equation (y2 – x3)dx – xydy = 0 (x eee 0) is : (where c is a constant of integration)
  1. A
    y2 + 2x3 + cx2 = 0
  2. B
    y2 + 2x2 + cx3 = 0
  3. C
    y2 – 2x + cx3 = 0
  4. D
    y2 – 2x3 + cx2 = 0
View written solutionFree

Correct answer: A

  1. Given differential equation
(y2−x3) dx−xy dy=0,(x≠0)(y^2-x^3)\,dx-xy\,dy=0, \qquad (x\neq 0)(y2−x3)dx−xydy=0,(x=0)

We need its general solution.

  1. Rewrite in differential form

Bring the dydydy term to the other side:

(y2−x3) dx=xy dy(y^2-x^3)\,dx=xy\,dy(y2−x3)dx=xydy

Now divide by dxdxdx:

y2−x3=xydydxy^2-x^3=xy\frac{dy}{dx}y2−x3=xydxdy​

So,

dydx=y2−x3xy\frac{dy}{dx}=\frac{y^2-x^3}{xy}dxdy​=xyy2−x3​

Simplify:

dydx=yx−x2y\frac{dy}{dx}=\frac{y}{x}-\frac{x^2}{y}dxdy​=xy​−yx2​
  1. Use substitution to reduce the equation

Since the terms suggest homogeneity in y2y^2y2 and x3x^3x3, let

v=y2v=y^2v=y2

Then,

dvdx=2ydydx\frac{dv}{dx}=2y\frac{dy}{dx}dxdv​=2ydxdy​

From the differential equation,

dydx=yx−x2y\frac{dy}{dx}=\frac{y}{x}-\frac{x^2}{y}dxdy​=xy​−yx2​

Multiply by 2y2y2y:

2ydydx=2y(yx−x2y)2y\frac{dy}{dx}=2y\left(\frac{y}{x}-\frac{x^2}{y}\right)2ydxdy​=2y(xy​−yx2​) dvdx=2y2x−2x2\frac{dv}{dx}=\frac{2y^2}{x}-2x^2dxdv​=x2y2​−2x2

Since v=y2v=y^2v=y2,

dvdx=2vx−2x2\frac{dv}{dx}=\frac{2v}{x}-2x^2dxdv​=x2v​−2x2

This is a linear differential equation:

dvdx−2xv=−2x2\frac{dv}{dx}-\frac{2}{x}v=-2x^2dxdv​−x2​v=−2x2
  1. Solve the linear differential equation

The integrating factor is

I.F.=e∫−2x dx=e−2ln⁡x=x−2\text{I.F.}=e^{\int -\frac{2}{x}\,dx}=e^{-2\ln x}=x^{-2}I.F.=e∫−x2​dx=e−2lnx=x−2

Multiply the equation by x−2x^{-2}x−2:

x−2dvdx−2xx−2v=−2x^{-2}\frac{dv}{dx}-\frac{2}{x}x^{-2}v=-2x−2dxdv​−x2​x−2v=−2

The left side becomes:

ddx(vx−2)=−2\frac{d}{dx}(v x^{-2})=-2dxd​(vx−2)=−2

Integrate both sides:

vx−2=−2x+cvx^{-2}=-2x+cvx−2=−2x+c

Multiply by x2x^2x2:

v=−2x3+cx2v=-2x^3+cx^2v=−2x3+cx2

Now substitute back v=y2v=y^2v=y2:

y2=−2x3+cx2y^2=-2x^3+cx^2y2=−2x3+cx2

Rearrange:

y2+2x3−cx2=0y^2+2x^3-cx^2=0y2+2x3−cx2=0

Since ccc is an arbitrary constant, replacing −c-c−c by ccc gives:

y2+2x3+cx2=0y^2+2x^3+cx^2=0y2+2x3+cx2=0
  1. Match with the options

The obtained general solution is

y2+2x3+cx2=0y^2+2x^3+cx^2=0y2+2x3+cx2=0

This matches Option A.

  1. Comparison with stored answer

Stored correct answer: A
Derived answer: A

So, the derived answer agrees with the stored answer.

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