# To solve: $\frac{dy}{dx}=\left(\frac{x+y+1}{x+y+3}\right)^{2}$.

I have to solve $\displaystyle \frac{dy}{dx}=\left(\frac{x+y+1}{x+y+3}\right)^{2}$.

i am taking $x+y=u.$ So i get $\displaystyle \frac{du}{dx}=\left(\frac{u+1}{u+3}\right)^{2}+1$. After this i dont know how to integrate this.

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This diff. equation can be written as $$\frac{dx}{dy}=\left(\frac{x+y+3}{x+y+1}\right)^2$$ Let $x+y+1=t\implies dx+dy=dt$ which converts your diff. equation to $$\frac{dt-dy}{dy}=\left(\frac{t+2}{t}\right)^2\implies \frac{dt}{dy}=1+\left(\frac{t+2}{t}\right)^2$$ which is variable separable and easy to integrate.
Are you sure that changing $dy/dx$ to $dx/dy$ wouldn't cause problems like necessity to inverse a function to obtain solution? –  Tigran Saluev Jul 10 '13 at 9:23