Please help me solve: 1. $\displaystyle y'(1+y'^{2})=ay''$ 2. $\displaystyle y^{3}y''=1$ 3. $\displaystyle xy''=y'ln\frac{y'}{x}$
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1 and 3 are reduced to first order by putting $\displaystyle t=y'.$
Originally Posted by sinichko Please help me solve: [snip] 2. $\displaystyle y^{3}y''=1$ [snip] $\displaystyle \frac{d^2y}{dx^2} = \frac{1}{y^3}$ is solved by first making the substitution $\displaystyle \frac{d^2 y}{dx^2} = \frac{d}{dy}\left[ \frac{v^2}{2}\right]$ where $\displaystyle v = \frac{dy}{dx}$ and solving for v.
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