Can we solve the following ODE via reduction of order? $y''+y'''sin(y)=0$.

Question

I am struggling with one problem in particular.
$y'' + y'''sin(y) = 0$ is the equation.
I am told to let $y' = z$ and do reduction of order to give:
$z'+z''sin(z)= 0$
But I am absolutely stuck. I don't know what a possible solution would be to do reduction of order. I thought about $Acos(y) = z$ as a solution.
Any advice?

doraemonpaul · Answer

Hint:
With reference to http://eqworld.ipmnet.ru/en/solutions/ode/ode0503.pdf,
Let $u=left(dfrac{dy}{dx}right)^2$ ,
Then $dfrac{du}{dx}=2dfrac{dy}{dx}dfrac{d^2y}{dx^2}$
$dfrac{du}{dy}dfrac{dy}{dx}=2dfrac{dy}{dx}dfrac{d^2y}{dx^2}$
$2dfrac{d^2y}{dx^2}=dfrac{du}{dy}$
$2dfrac{d^3y}{dx^3}=dfrac{d}{dx}left(dfrac{du}{dy}right)=dfrac{d}{dy}left(dfrac{du}{dy}right)dfrac{dy}{dx}=pmsqrt udfrac{d^2u}{dy^2}$
$thereforedfrac{du}{dy}pmsqrt udfrac{d^2u}{dy^2}sin y=0$

Can we solve the following ODE via reduction of order? $y''+y'''sin(y)=0$.

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