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Question Number 92040 by mhmd last updated on 04/May/20

$$y^{{}^{″}}+{\left(y^{\prime }\right)}^2+y=0y\left(0\right)=-\frac{1}{2},y^{\prime }\left(0\right)=-1$$$$helpmesir$$

Answered by mr W last updated on 04/May/20

$$letu=y^{\prime }=\frac{dy}{dx}$$$$y^{″}=\frac{du}{dx}=\frac{du}{dy}\times \frac{dy}{dx}=u\frac{du}{dy}$$$$u\frac{du}{dy}+u^2+y=0$$$$\frac{1}{2}\frac{d\left(u^2\right)}{dy}+u^2+y=0$$$$letU=u^2$$$$\frac{1}{2}\frac{dU}{dy}+U+y=0$$$$\frac{dU}{dy}+2U=-2y$$$$U=\frac{-\int 2{ye}^{\int 2dy}dy+C}{e^{\int 2fy}}$$$$U=\frac{-\int 2{ye}^{2y}dy+C}{e^{2y}}$$$$U=\frac{-{ye}^{2y}+\int e^{2y}dy+C}{e^{2y}}$$$$U=\frac{-{ye}^{2y}+\frac{1}{2}{e}^{2y}+C}{e^{2y}}$$$$U=\frac{1-2y+C_1{e}^{-2y}}{2}=u^2={\left(\frac{dy}{dx}\right)}^2$$$$\frac{dy}{dx}=\pm \frac{\sqrt{1-2y+C_1{e}^{-2y}}}{\sqrt{2}}$$$$-1=\pm \frac{\sqrt{2+C_{1}e}}{\sqrt{2}}$$$$\Rightarrow C_1=0$$$$\frac{dy}{\sqrt{1-2y}}=-\frac{dx}{\sqrt{2}}$$$$x=-\sqrt{2}\int \frac{dy}{\sqrt{1-2y}}+C_2$$$$x=\sqrt{2(1-2y)}+C_2$$$$0=\sqrt{2(1+1)}+C_2$$$$\Rightarrow C_2=-2$$$$\Rightarrow x=\sqrt{2(1-2y)}-2$$$$or$$$$\Rightarrow y=\frac{1}{2}-\frac{1}{4}{(x+2)}^2$$

Commented by mr W last updated on 04/May/20

$$check:$$$$y\left(0\right)=\frac{1}{2}-1=-\frac{1}{2}$$$$y^{\prime }=-\frac{x+2}{2}=-1-\frac{x}{2}$$$$y^{\prime }\left(0\right)=-1$$$$y^{″}=-\frac{1}{2}$$$$y^{″}+{\left(y^{\prime }\right)}^2+y=-\frac{1}{2}+\frac{{(x+2)}^2}{4}+\frac{1}{2}-\frac{{(x+2)}^2}{4}=0$$

Commented by niroj last updated on 04/May/20

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