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Differential Equations question

2020 · 2 Sep · Shift 2 · Q30
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  5. /2020 · 2 Sep · Shift 2 · Q30

Differential Equations question

2020 · 2 Sep · Shift 2 · Q30

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation, 2x2dy= (2xy + y2)dx, then f(12)f\left( {{1 \over 2}} \right)f(21​) is equal to :
  1. A
    11−log⁡e2{1 \over {1 - {{\log }_e}2}}1−loge​21​
  2. B
    11+log⁡e2{1 \over {1 + {{\log }_e}2}}1+loge​21​
  3. C
    −11+log⁡e2{{ - 1} \over {1 + {{\log }_e}2}}1+loge​2−1​
  4. D
    1+log⁡e2{1 + {{\log }_e}2}1+loge​2
View written solutionFree

Correct answer: B

  1. Write the differential equation in standard form

Given 2x2 dy=(2xy+y2) dx2x^2\,dy=(2xy+y^2)\,dx2x2dy=(2xy+y2)dx

So, dydx=2xy+y22x2\frac{dy}{dx}=\frac{2xy+y^2}{2x^2}dxdy​=2x22xy+y2​

Simplify: dydx=yx+y22x2\frac{dy}{dx}=\frac{y}{x}+\frac{y^2}{2x^2}dxdy​=xy​+2x2y2​

This is a homogeneous differential equation.

  1. Use the substitution

Let y=vxy=vxy=vx Then dydx=v+xdvdx\frac{dy}{dx}=v+x\frac{dv}{dx}dxdy​=v+xdxdv​

Substitute into the differential equation: v+xdvdx=v+v22v+x\frac{dv}{dx}=v+\frac{v^2}{2}v+xdxdv​=v+2v2​

Therefore, xdvdx=v22x\frac{dv}{dx}=\frac{v^2}{2}xdxdv​=2v2​

So, dvv2=dx2x\frac{dv}{v^2}=\frac{dx}{2x}v2dv​=2xdx​

  1. Integrate both sides

∫v−2 dv=∫12x dx\int v^{-2}\,dv=\int \frac{1}{2x}\,dx∫v−2dv=∫2x1​dx

−1v=12ln⁡∣x∣+C-\frac{1}{v}=\frac{1}{2}\ln|x|+C−v1​=21​ln∣x∣+C

Multiply by −1-1−1: 1v=C−12ln⁡∣x∣\frac{1}{v}=C-\frac{1}{2}\ln|x|v1​=C−21​ln∣x∣

Since v=yxv=\frac{y}{x}v=xy​, 1y/x=xy=C−12ln⁡∣x∣\frac{1}{y/x}=\frac{x}{y}=C-\frac{1}{2}\ln|x|y/x1​=yx​=C−21​ln∣x∣

Hence, xy=C−12ln⁡∣x∣\frac{x}{y}=C-\frac{1}{2}\ln|x|yx​=C−21​ln∣x∣

So, y=xC−12ln⁡∣x∣y=\frac{x}{C-\frac{1}{2}\ln|x|}y=C−21​ln∣x∣x​

  1. Use the point (1,2)(1,2)(1,2) to find CCC

Given that the curve passes through (1,2)(1,2)(1,2): 2=1C−12ln⁡12=\frac{1}{C-\frac{1}{2}\ln 1}2=C−21​ln11​

Since ln⁡1=0\ln 1=0ln1=0, 2=1C2=\frac{1}{C}2=C1​ Rightarrow C=12C=\frac{1}{2}C=21​

Thus,

=\frac{2x}{1-\ln|x|}$$ --- 5. **Find $f\left(\frac12\right)$** $$f\left(\frac12\right)=\frac{2\cdot \frac12}{1-\ln\left(\frac12\right)}$$ Since $$\ln\left(\frac12\right)=-\ln 2$$ we get $$f\left(\frac12\right)=\frac{1}{1-(-\ln 2)} =\frac{1}{1+\ln 2}$$ --- 6. **Compare with the options** $$f\left(\frac12\right)=\frac{1}{1+\ln 2}$$ This matches **Option B**.
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