Help me with this

$\displaystyle y'' = 18\sin ^3y\cos y$, $\displaystyle y(1) =\frac{\pi}{2}$, $\displaystyle y'(1) = 3$.

I tried to solve it, but I got non-elementary integral (Crying)

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- Jan 12th 2010, 08:01 PMMentThe second order ODE
Help me with this

$\displaystyle y'' = 18\sin ^3y\cos y$, $\displaystyle y(1) =\frac{\pi}{2}$, $\displaystyle y'(1) = 3$.

I tried to solve it, but I got non-elementary integral (Crying) - Jan 13th 2010, 03:26 AMmr fantastic
- Jan 13th 2010, 04:15 AMMent
Please give more tips.

Here's my solution

$\displaystyle y'' = 18{\sin ^3}y\cos y, y(1) = \frac{\pi}{2},y'(1) = 3.$

$\displaystyle y' = p(y) \Rightarrow y'' = p'(y)p(y)$

$\displaystyle p'p = 18\sin^3y\cos y$

$\displaystyle \int pdp = 18\int \sin^3y\cos ydy$

$\displaystyle \frac{p^2}{2} = \frac{9}{2}\sin^4y + C_1 \Rightarrow (y')^2 = 9\sin^4y + C_1$

$\displaystyle y' = \pm \sqrt {9\sin^4y + C_1}$

$\displaystyle x = \pm \int \frac{dy}{\sqrt{9\sin^4y + C_1}} $

I got non-elementary integral. - Jan 13th 2010, 04:23 AMHallsofIvy
I guess, then, the question is why you would

**expect**to get a solution in terms of an elementary integral! That's a**very**non-linear equation and non-linear differential equations tend to have non-elementary solutions. Something as simple (comparatively) as y"= sin(y) has solutions that can only be written in terms of elliptic integrals. - Jan 13th 2010, 10:05 AMDeMath
You do not necessarily find a general solution to solve your problem of Cauchy.

You have the initial conditions in the symmetrical form: $\displaystyle x=1\ \leftrightarrow\ y={\pi\over2}\ \leftrightarrow\ y'=3$

Also you got $\displaystyle \left(y'\right)^2 = 9\sin^4y + C_1$. Now find constant $\displaystyle C_1$:

$\displaystyle \left\{\begin{gathered}\left(y'\right)^2 = 9\sin^4y + C_1, \hfill \\ y(1) = \frac{\pi}{2} ~\wedge~ y'\left( 1 \right) = 3; \hfill \\ \end{gathered} \right. ~ \Rightarrow ~ 3^2 = 9 \sin^4\frac{\pi}{2} + C_1 ~ \Leftrightarrow ~ C_1 = 0.$

So you have

$\displaystyle \left(y'\right)^2 = 9\sin^4y ~ \Leftrightarrow ~ y' = 3\sin^2y ~ \Rightarrow ~ x = \frac{1}{3}\int \frac{dx}{\sin^2y} = C_2 - \frac{1}{3}\cot y.$

Now find the constant $\displaystyle C_2$:

$\displaystyle \left\{ \begin{gathered}x = C_2 - \frac{1}{3}\cot y, \hfill \\ y\left( 1 \right) = \frac{\pi}{2}; \hfill \\ \end{gathered} \right. ~ \Rightarrow ~ 1 = C_2 - \frac{1}{3}\cot\frac{\pi}{2} ~ \Leftrightarrow ~ C_2 = 1.$

Finally you have:

$\displaystyle x = 1 - \frac{1}{3}\cot y.$