The general solution of the differential equation \; dy/dx = \; x+y/x-y is [AP EAPCET 22 May 2025 S2]

Correct Answer is : tan⁡−1(yx)=log⁡(cx2+y2)^{-1} ≤ft( y/x ) = ≤ft( c {x²+y² )tan−1(xy)=log(cx2+y2)

^{-1} ≤ft( y/x ) = ≤ft( c x {x²+y² )

^{-1} ≤ft( y/x ) = ≤ft( c {x²+y² )

Correct Answer is : tan⁡−1(yx)=log⁡(cx2+y2)^{-1} ≤ft( y/x ) = ≤ft( c {x²+y² )tan−1(xy)=log(cx2+y2)

Test Index

Differential Equations

Section: Mathematics

Question : 12 of 48

Marks: +1, -0

\; \frac{dy}{dx} = \; \frac{x+y}{x-y}

Solution:

\;\; \frac{dy}{dx} = \; \frac{x+y}{x-y}

\Rightarrow y=v x \Rightarrow v=\; \frac{y}{x}

\Rightarrow \;\; \frac{dy}{dx} = v + x \; \frac{dv}{dx}

\Rightarrow v + x \; \frac{dv}{dx} = \; \frac{x+v x}{x-v x}

\Rightarrow v + x \; \frac{dv}{dx} = \; \frac{1+v}{1-v}

\Rightarrow x \; \frac{dv}{dx} = \; \frac{1+v}{1-v} - v

\Rightarrow = \; \frac{1+v-v+v^2}{1-v}

\Rightarrow x \; \frac{dv}{dx} = \; \frac{1+v^2}{1-v}

\Rightarrow \; \frac{1-v}{1+v^2} dv = \; \frac{1}{x} dx

Integrating both sides, we get

\;\Rightarrow \;\; \int \frac{1-v}{1+v^2} dv = \int \frac{1}{x} dx

\;\Rightarrow \;\; \log x = \int \frac{1}{1+v^2} dv - \int \frac{v}{1+v^2} dv

\;\Rightarrow \;\; \log x = \tan^{-1} v - \frac{1}{2} \int \frac{1}{t} dt

where,

1+v^2=t

2 v dv = dt

\; v dv = \; \frac{1}{2} dt

\;\Rightarrow \log x = \tan^{-1} v - \frac{1}{2} \log |1+v^2| + C

\;\Rightarrow \log x = \tan^{-1} \frac{y}{x} - \frac{1}{2} \log \left| 1+\frac{y^2}{x^2} \right| + C

\;\Rightarrow \log x = \tan^{-1} \frac{y}{x} - \frac{1}{2} \log \left| \frac{y^2+x^2}{x^2} \right| + C

\;\Rightarrow \tan^{-1} \left( \frac{y}{x} \right) = \log \left( c \sqrt{x^2+y^2} \right)

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