General solution of $yy′′+2y'''=0$ How do you derive the general solution of $yy''+2y'''= 0$?
please help me to derive solution thanks a lot

NB: This is a general way to reduce the order of the equation. This doesn't solve your question as is, but rather gives you a starting point.
In this equation, the variable $t$ does not appear, hence one can substitute: $$y'=p(y), \hspace{7pt}y''=p'(y)\cdot y'=p'p,\hspace{7pt}y'''=p''\cdot y'\cdot p+p'\cdot p'\cdot y'=p''p^2+(p')^2p$$
(where $y'=\frac{dy}{dt}$, while $p'=\frac{dp}{dy}$). 
Then your equation turns into 
$$yp'p+2p''p^2+2(p')^2p=0 \hspace{7pt}\Rightarrow\hspace{7pt} yp'+2p''p+2(p')^2=0$$
Where the derivatives are w.r.t $y$ - this gives you an equation of the second order.
