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Question Number 123282 by zarminaawan last updated on 24/Nov/20

$$y-x\frac{dy}{dx}=a(y\times y+\frac{dy}{dx})$$

Answered by Dwaipayan Shikari last updated on 24/Nov/20

$$y-x\frac{dy}{dx}={ay}^2+a\frac{dy}{dx}$$$$\frac{dy}{dx}(x+a)=y(1-ay)$$$$\int \frac{dy}{y(1-ay)}=\int \frac{dx}{(x+a)}$$$$\Rightarrow \int \frac{1}{y}-\frac{a}{ay-1}=log(x+a)+C$$$$\Rightarrow logy-log(y-\frac{1}{a})=log(x+a)+C$$$$\Rightarrow log\left(\frac{y}{y-\frac{1}{a}}\right)=log(C_{1}x+Ca)$$$$\Rightarrow \frac{y}{y-\frac{1}{a}}=C_{1}x+Ca\Rightarrow \frac{y-\frac{1}{a}}{y}=\frac{1}{C_1(x+a)}\Rightarrow \frac{1}{ay}=1-\frac{1}{C_1(x+a)}$$$$y=\frac{C_1(x+a)}{a(C_1(x+a)-1)}$$

Commented by zarminaawan last updated on 24/Nov/20

$$thankyousir$$

Answered by TANMAY PANACEA last updated on 24/Nov/20

$$y-x\frac{dy}{dx}={ay}^2+a\frac{dy}{dx}$$$$ydx-xdy={ay}^{2}dx+ady$$$$\frac{ydx-xdy}{y^2}=adx+a\frac{dy}{y^2}$$$$d\left(\frac{x}{y}\right)=adx+{ay}^{-2}dy$$$$intregating$$$$\frac{y}{x}=ax+a\times \frac{y^{-1}}{-1}+c$$$$\frac{y}{x}=ax-\frac{a}{y}+C$$

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