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Question Number 181330 by Mastermind last updated on 24/Nov/22

$$\mathrm{Solve}:$$$$\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{x}(\mathrm{y}+1)=0,\mathrm{y}\left(0\right)=2$$

Answered by FelipeLz last updated on 24/Nov/22

$$\frac{dy}{dx}+2x(y+1)=0$$$$\frac{dy}{dx}+2xy=-2x$$$$\frac{dy}{dx}{e}^{x^2}+2{xe}^{x^2}y=-2{xe}^{x^2}$$$$\frac{d}{dx}\left[{ye}^{x^2}\right]=-2{xe}^{x^2}$$$${ye}^{x^2}=-\int 2{xe}^{x^2}dx$$$${ye}^{x^2}=-e^{x^2}+c$$$$y\left(x\right)={ce}^{-x^2}-1$$$$y\left(0\right)=2\to {ce}^{-0^2}-1=2\Rightarrow c=3$$$$y\left(x\right)=3e^{-x^2}-1$$$$\bullet$$$$\frac{dy}{dx}+2x(y+1)=0$$$$\frac{dy}{dx}=-2x(y+1)$$$$\frac{1}{y+1}dy=-2xdx$$$$\int \frac{1}{y+1}dy=-\int 2xdx$$$$\ln \mid y+1\mid +c_1=-x^2+c_2$$$$\ln \mid y+1\mid =-x^2+c_2-c_1$$$$c_2-c_1=c_3\to \ln \mid y+1\mid =-x^2+c_3$$$$y+1=e^{-x^2+c_3}$$$$e^{c_3}=C\to y+1={Ce}^{-x^2}$$$$y\left(x\right)={Ce}^{-x^2}-1$$$$y\left(0\right)=2\to {Ce}^{-0^2}-1=2\Rightarrow C=3$$$$y\left(x\right)=3e^{-x^2}-1$$

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