# The second order ODE

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• Jan 12th 2010, 09:01 PM
Ment
The second order ODE
Help me with this

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

I tried to solve it, but I got non-elementary integral (Crying)
• Jan 13th 2010, 04:26 AM
mr fantastic
Quote:

Originally Posted by Ment
Help me with this

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

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

Note that $\frac{d^2y}{dt^2} = f(y) \Rightarrow v \frac{dv}{dy} = f(y)$ where $v = \frac{dy}{dt}$.

So solve for v (note that the DE is seperable) and then use your solution for v to solve for y.
• Jan 13th 2010, 05:15 AM
Ment
Quote:

Originally Posted by mr fantastic
Note that $\frac{d^2y}{dt^2} = f(y) \Rightarrow v \frac{dv}{dy} = f(y)$ where $v = \frac{dy}{dt}$.

So solve for v (note that the DE is seperable) and then use your solution for v to solve for y.

Please give more tips.

Here's my solution

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

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

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

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

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

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

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

I got non-elementary integral.
• Jan 13th 2010, 05:23 AM
HallsofIvy
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, 11:05 AM
DeMath
Quote:

Originally Posted by Ment

Here's my solution

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

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

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

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

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

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

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

I got non-elementary integral.

You do not necessarily find a general solution to solve your problem of Cauchy.

You have the initial conditions in the symmetrical form: $x=1\ \leftrightarrow\ y={\pi\over2}\ \leftrightarrow\ y'=3$
Also you got $\left(y'\right)^2 = 9\sin^4y + C_1$. Now find constant $C_1$:

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

So you have

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

Now find the constant $C_2$:

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

Finally you have:

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