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Question Number 167838 by cortano1 last updated on 27/Mar/22

$$\frac{dy}{dx}=8x+4y+{(2x+y-1)}^2$$$$y=?$$

Answered by mr W last updated on 27/Mar/22

$$letu=2x+y-1$$$$\frac{du}{dx}=2+\frac{dy}{dx}$$$$\frac{du}{dx}-2=4u+4+u^2$$$$\frac{du}{u^2+4u+6}=dx$$$$\frac{du}{{(u+2)}^2+2}=dx$$$$\int \frac{d\left(\frac{u+2}{\sqrt{2}}\right)}{{\left(\frac{u+2}{\sqrt{2}}\right)}^2+1}=\sqrt{2}\int dx$$$${\tan }^{-1}\frac{u+2}{\sqrt{2}}=\sqrt{2}x+C$$$$u+2=\sqrt{2}\tan (\sqrt{2}x+C)$$$$2x+y+1=\sqrt{2}\tan (\sqrt{2}x+C)$$$$\Rightarrow y=\sqrt{2}\tan (\sqrt{2}x+C)-2x-1$$

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