# Solve y''=−2x(y)

Solve $y=-2x\left(y\right)$
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Khribechy

Given that y''=−2x(y)
${y}^{″}=-2x\left({y}^{\prime }{\right)}^{2}->{y}^{″}/\left({y}^{\prime }{\right)}^{2}=-2x->d\left({y}^{\prime }\right)/\left({y}^{\prime }{\right)}^{2}=-2x->\int \left(d\left({y}^{\prime }\right)/\left({y}^{\prime }{\right)}^{2}\right)=-\int 2x-c0$

where ${c}_{0}$ is a constant $\to \left(1/{y}^{\prime }\right)=-{x}^{2}-{c}_{0}\to 1/{y}^{\prime }={x}^{2}+{c}_{0}\to {y}^{\prime }=1/\left(\left({x}^{2}\right)+{c}_{0}\right)\to \int {y}^{\prime }=\int \left(1/\left({x}^{2}+{c}_{1}^{2}+{c}_{2}\right)$

where ${c}_{0}={c}_{1}^{2}$ and ${c}_{1}$ are constant $\to y=1/{c}_{1}{\mathrm{tan}}^{-1}\left(x/{c}_{1}\right)+{c}_{2}$

Hence the solution of the given Differential equations is $y=\frac{1}{c}{\mathrm{tan}}^{-1}\left(\frac{x}{c}\right)+{c}_{2}$