I am a 12th grader in India . I am currently in my Differential Equations Course and I am a beginner. Can anybody help me in solving this Differential Equation. I could nowhere find a solution on web .

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    $\begingroup$ Hint: put $$X=x+2,Y=y-2$$ becuse $x+y=x+2+y-2$ $\endgroup$ Dec 4, 2020 at 8:17
  • $\begingroup$ That Solved the question . I tried all sorts of Manipulations I could imagine and did not think about doing something so simple. Thanks $\endgroup$
    – Harsh
    Dec 4, 2020 at 8:21
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    $\begingroup$ i am glad you are tried,ideally you could self answer this helps other people who may see the question and get their query answered $\endgroup$ Dec 4, 2020 at 8:22
  • $\begingroup$ @Harsh Welcome to MSE. Learn about the site and its features and how to ask. $\endgroup$
    – user730361
    Dec 4, 2020 at 8:30

1 Answer 1


$$\frac{dy}{dx}=\frac{(x+y)^2}{(x+2)(y-2)}$$ Let $x-2=X, y-2=Y$, then $$\frac{dY}{dX}=\frac{(X+Y)^2}{XY}$$ Let $$Y=VX \implies \frac{dY}{dx}=X\frac{dV}{dX}+V$$ $$\implies X\frac{dV}{dX}+V= \frac{(1+V)^2}{V} \implies X\frac{dV}{dX}=\frac{(1+V)^2}{V}-V \implies \int \frac{VdV} {1+2V}= \int \frac{dX}{X}$$
$$\implies V=2\ln[CX(1+2V)^{1/4}], V=Y/X, X=x+2. Y=y-2$$ $$


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