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Differential Equations question

2020 · 3 Sep · Shift 2 · Q32
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  5. /2020 · 3 Sep · Shift 2 · Q32

Differential Equations question

2020 · 3 Sep · Shift 2 · Q32

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If x3dy + xy dx = x2dy + 2y dx; y(2) = e and x > 1, then y(4) is equal to :
  1. A
    e2{{\sqrt e } \over 2}2e​​
  2. B
    12+e{1 \over 2} + \sqrt e21​+e​
  3. C
    32+e{3 \over 2} + \sqrt e23​+e​
  4. D
    32e{3 \over 2}\sqrt e23​e​
View written solutionFree

Correct answer: D

  1. Write the differential equation clearly

Given: x3 dy+xy dx=x2 dy+2y dxx^3\,dy + xy\,dx = x^2\,dy + 2y\,dxx3dy+xydx=x2dy+2ydx with initial condition y(2)=e,x>1.y(2)=e, \qquad x>1.y(2)=e,x>1.

We need to find y(4)y(4)y(4).

  1. Rearrange the equation

Bring like terms together: x3dy−x2dy=2y dx−xy dxx^3dy - x^2dy = 2y\,dx - xy\,dxx3dy−x2dy=2ydx−xydx

Factor: (x3−x2)dy=y(2−x)dx(x^3-x^2)dy = y(2-x)dx(x3−x2)dy=y(2−x)dx

x2(x−1)dy=−y(x−2)dxx^2(x-1)dy = -y(x-2)dxx2(x−1)dy=−y(x−2)dx

So, dyy=−x−2x2(x−1) dx\frac{dy}{y} = -\frac{x-2}{x^2(x-1)}\,dxydy​=−x2(x−1)x−2​dx

  1. Simplify using partial fractions

We decompose: x−2x2(x−1)=Ax+Bx2+Cx−1\frac{x-2}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}x2(x−1)x−2​=xA​+x2B​+x−1C​

Thus, x−2=Ax(x−1)+B(x−1)+Cx2x-2 = A x(x-1) + B(x-1) + Cx^2x−2=Ax(x−1)+B(x−1)+Cx2

Expanding: x−2=(A+C)x2+(−A+B)x−Bx-2 = (A+C)x^2 + (-A+B)x - Bx−2=(A+C)x2+(−A+B)x−B

Compare coefficients:

  • Constant term: −B=−2⇒B=2-B=-2 \Rightarrow B=2−B=−2⇒B=2
  • Coefficient of xxx: −A+B=1⇒−A+2=1⇒A=1-A+B=1 \Rightarrow -A+2=1 \Rightarrow A=1−A+B=1⇒−A+2=1⇒A=1
  • Coefficient of x2x^2x2: A+C=0⇒1+C=0⇒C=−1A+C=0 \Rightarrow 1+C=0 \Rightarrow C=-1A+C=0⇒1+C=0⇒C=−1

Hence, x−2x2(x−1)=1x+2x2−1x−1\frac{x-2}{x^2(x-1)}=\frac{1}{x}+\frac{2}{x^2}-\frac{1}{x-1}x2(x−1)x−2​=x1​+x22​−x−11​

Therefore, dyy=−(1x+2x2−1x−1)dx\frac{dy}{y}= -\left(\frac1x+\frac{2}{x^2}-\frac1{x-1}\right)dxydy​=−(x1​+x22​−x−11​)dx

  1. Integrate both sides

∫dyy=−∫(1x+2x2−1x−1)dx\int \frac{dy}{y} = -\int \left(\frac1x+\frac{2}{x^2}-\frac1{x-1}\right)dx∫ydy​=−∫(x1​+x22​−x−11​)dx

ln⁡y=−ln⁡x+2x+ln⁡(x−1)+C\ln y = -\ln x + \frac{2}{x} + \ln(x-1) + Clny=−lnx+x2​+ln(x−1)+C

So, ln⁡y=ln⁡(x−1x)+2x+C\ln y = \ln\left(\frac{x-1}{x}\right) + \frac{2}{x} + Clny=ln(xx−1​)+x2​+C

Exponentiating, y=C x−1xe2/xy = C\,\frac{x-1}{x} e^{2/x}y=Cxx−1​e2/x

  1. Use the initial condition y(2)=ey(2)=ey(2)=e

Substitute x=2,y=ex=2, y=ex=2,y=e: e=C⋅2−12⋅e2/2e = C\cdot \frac{2-1}{2} \cdot e^{2/2}e=C⋅22−1​⋅e2/2

e=C⋅12⋅ee = C\cdot \frac12 \cdot ee=C⋅21​⋅e

So, C=2C=2C=2

Thus, y=2⋅x−1xe2/xy = 2\cdot \frac{x-1}{x} e^{2/x}y=2⋅xx−1​e2/x

  1. Find y(4)y(4)y(4)

y(4)=2⋅4−14e2/4y(4)=2\cdot \frac{4-1}{4} e^{2/4}y(4)=2⋅44−1​e2/4

y(4)=2⋅34⋅e1/2y(4)=2\cdot \frac34 \cdot e^{1/2}y(4)=2⋅43​⋅e1/2

y(4)=32ey(4)=\frac32\sqrt ey(4)=23​e​

  1. Compare with the given options

y(4)=32ey(4)=\frac32\sqrt ey(4)=23​e​

This matches Option D.

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