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GATE 2018 EC – Question 44

Engineering Mathematics · Differential Equations · 2 marks · Multiple choice

A curve passes through the point $(x=1,\ y=0)$ and satisfies the differential equation

$$\frac{dy}{dx}=\frac{x^2+y^2}{2y}+\frac yx.$$

The equation that describes the curve is

  1. $\ln\left(1+\dfrac{y^2}{x^2}\right)=x-1$
  2. $\dfrac12\ln\left(1+\dfrac{y^2}{x^2}\right)=x-1$
  3. $\ln\left(1+\dfrac yx\right)=x-1$
  4. $\dfrac12\ln\left(1+\dfrac yx\right)=x-1$

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Correct answer: (A) $\ln\left(1+\dfrac{y^2}{x^2}\right)=x-1$

Explanation

Let $w=y^2$, so $\frac{dw}{dx}=2y\frac{dy}{dx}=x^2+w+\frac{2w}{x}$. This is linear: $w'-\left(1+\frac2x\right)w=x^2$. With integrating factor $\frac{e^{-x}}{x^2}$ we get $\frac{d}{dx}\left(\frac{we^{-x}}{x^2}\right)=e^{-x}$, so $\frac{w}{x^2}e^{-x}=-e^{-x}+C$. The point $(1,0)$ gives $C=e^{-1}$, so $1+\frac{y^2}{x^2}=e^{x-1}$, i.e. $\ln\left(1+\frac{y^2}{x^2}\right)=x-1$.