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Differential Equations question

2008 · Shift 0 · Q35
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Differential Equations question

2008 · Shift 0 · Q35

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
The solution of the differential equation dydx=x+yx{{dy} \over {dx}} = {{x + y} \over x}dxdy​=xx+y​ satisfying the condition y(1)=1y(1)=1y(1)=1 is :
  1. A
    y=ln⁡x+xy = \ln x + xy=lnx+x
  2. B
    y=xln⁡x+x2y = x\ln x + {x^2}y=xlnx+x2
  3. C
    y=xe(x−1) y = x{e^{\left( {x - 1} \right)}}\,y=xe(x−1)
  4. D
    y=x ln⁡x+xy = x\,\ln x + xy=xlnx+x
View written solutionFree

Correct answer: D

  1. Given differential equation

dydx=x+yx\frac{dy}{dx} = \frac{x+y}{x}dxdy​=xx+y​

Simplify the right-hand side:

dydx=1+yx\frac{dy}{dx} = 1 + \frac{y}{x}dxdy​=1+xy​

Rearrange:

dydx−1xy=1\frac{dy}{dx} - \frac{1}{x}y = 1dxdy​−x1​y=1

This is a linear differential equation of the form

dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)dxdy​+P(x)y=Q(x)

with

P(x)=−1x,Q(x)=1P(x) = -\frac{1}{x}, \qquad Q(x)=1P(x)=−x1​,Q(x)=1

  1. Find the integrating factor

I.F.=e∫P(x) dx=e∫−1x dx=e−ln⁡x=1x\text{I.F.} = e^{\int P(x)\,dx} = e^{\int -\frac{1}{x}\,dx} = e^{-\ln x} = \frac{1}{x}I.F.=e∫P(x)dx=e∫−x1​dx=e−lnx=x1​

  1. Multiply the equation by the integrating factor

1xdydx−yx2=1x\frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2} = \frac{1}{x}x1​dxdy​−x2y​=x1​

The left side is:

ddx(yx)\frac{d}{dx}\left(\frac{y}{x}\right)dxd​(xy​)

So,

ddx(yx)=1x\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{1}{x}dxd​(xy​)=x1​

  1. Integrate both sides

yx=∫1x dx=ln⁡x+C\frac{y}{x} = \int \frac{1}{x}\,dx = \ln x + Cxy​=∫x1​dx=lnx+C

Hence,

y=x(ln⁡x+C)=xln⁡x+Cxy = x(\ln x + C) = x\ln x + Cxy=x(lnx+C)=xlnx+Cx

  1. Use the initial condition

Given y(1)=1y(1)=1y(1)=1:

1=1⋅ln⁡1+C(1)1 = 1\cdot \ln 1 + C(1)1=1⋅ln1+C(1)

Since ln⁡1=0\ln 1 = 0ln1=0,

1=C1 = C1=C

Therefore,

y=xln⁡x+xy = x\ln x + xy=xlnx+x

  1. Check with options
  • A: y=ln⁡x+xy=\ln x + xy=lnx+x ❌
  • B: y=xln⁡x+x2y=x\ln x + x^2y=xlnx+x2 ❌
  • C: y=xex−1y=xe^{x-1}y=xex−1 ❌
  • D: y=xln⁡x+xy=x\ln x + xy=xlnx+x ✅

So the correct option is D.

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