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Question Number 181618 by Mastermind last updated on 27/Nov/22

$$\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{xy}={\mathrm{x}}^2\mathrm{y}\left(0\right)=3$$$$$$$$\mathrm{Solve}$$$$$$$$.$$

Answered by hmr last updated on 27/Nov/22

$$multiplybothsidesofeq.$$$$toafunctionlike\mu \left(x\right).$$$$\mu \left(x\right)\frac{dy}{dx}+2\mu \left(x\right)xy=\mu \left(x\right)x^2$$$$$$$$trytoformtheLHSlikethis:$$$$\frac{d}{dx}\left(\mu \left(x\right)y\right)=\frac{d\mu }{dx}y+\mu \left(x\right)\frac{dy}{dx}$$$$$$$$soblueonesshouldbeequal.$$$$\frac{d\mu }{dx}y=2\mu \left(x\right)xy$$$$\frac{d\mu /dx}{\mu \left(x\right)}=2x$$$$\frac{d}{dx}(ln\mid \mu \left(x\right)\mid )=2x$$$$ln\mid \mu \left(x\right)\mid =x^2+c$$$$\mu \left(x\right)=\pm e^c{e}^{x^2}=C{e}^{x^2}$$$$C=1\to \mu \left(x\right)=e^{x^2}$$$$$$$$replace\mu \left(x\right)inequation.$$$$\frac{d}{dx}\left(e^{x^2}y\right)=e^{x^2}{x}^2$$$$e^{x^2}y=\int e^{x^2}{x}^{2}dx$$$$y=\frac{\int e^{x^2}{x}^{2}dx+c}{e^{x^2}}$$$$$$

Answered by ali009 last updated on 28/Nov/22

$$usingtheruleoflinearfirstordereq$$$$\frac{dy}{dx}+p\left(x\right)y=Q\left(x\right)$$$$y=\frac{1}{D\left(x\right)}\int D\left(x\right)Q\left(x\right)dx$$$$D\left(x\right)=e^{\int p\left(x\right)dx}=e^{\int 2xdx}=e^{x^2}$$$$y=\frac{1}{e^{x^2}}(\int e^{x^2}{x}^{2}dx+c)$$

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