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Question Number 202591 by Calculusboy last updated on 30/Dec/23

Answered by mr W last updated on 30/Dec/23

$$\frac{dy}{dx}+1+\frac{y}{x}=0$$$$lety=xt$$$$\frac{dy}{dx}=t+x\frac{dt}{dx}$$$$t+x\frac{dt}{dx}+1+t=0$$$$\frac{dt}{2t+1}=-\frac{dx}{x}$$$$\int \frac{dt}{2t+1}=-\int \frac{dx}{x}$$$$\frac{1}{2}\ln (2t+1)=-\ln x+C_1$$$$2t+1=\frac{C^2}{x^2}$$$$\frac{2y}{x}+1=\frac{C^2}{x^2}$$$$\Rightarrow y=\frac{1}{2}(\frac{C^2}{x}-x)=\frac{C^2-x^2}{2x}$$

Commented by Calculusboy last updated on 30/Dec/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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