The solution of the equation (x²+x y) d y=(x²+y²) d x is

Correct Answer is : log⁡x=2log⁡(x−y)+ yx+C x=2 (x-y)+ \; y/x+Clogx=2log(x−y)+xy+C

Correct Answer is : log⁡x=2log⁡(x−y)+ yx+C x=2 (x-y)+ \; y/x+Clogx=2log(x−y)+xy+C

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MHT CET 2021 Math Solved Paper

Question : 94 of 100

Marks: +1, -0

(x^{2}+x y) d y=(x^{2}+y^{2}) d x

Solution:

Given,

(x^{2}+x y) d y=(x^{2}+y^{2}) d x

\Rightarrow \; \; \; \frac{d y}{d x}= \; \frac{x^{2}+y^{2}}{x^{2}+x y}

Put

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

\Rightarrow \; \; v+x \; \frac{d v}{d x}= \; \frac{x^{2}+v^{2} x^{2}}{x^{2}+x^{2} v}

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

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

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

\Rightarrow \; \; d v \left( \; \frac{1+v}{1-v} \right)= \; \frac{d x}{x} \Rightarrow d v \left( -1+ \; \frac{2}{1-v} \right)= \; \frac{d x}{x}

On integrating both sides, we get

\; \; -v-2 \log (1-v)=\log x+C

\Rightarrow \; \; \; \; \; \frac{y}{x}-2 \log \left(1- \; \frac{y}{x}\right)=\log x+C

\Rightarrow \; \; \; \frac{y}{x}-2 \log (x-y)+2 \log (x-y)+C=\log x+C

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