# Solving a second order ODE substituting $y'=v$

I was given this problem

$$y'y'' = 1$$

and told to solve it using the method in which

$y' = v$ and $y'' = v \frac{dv}{dy}$.

I had previously solved it using the method in which you substitute

$v = y'$ and $v' = y''$ getting $$y=\frac{1}{3}(2x + c_1)^\frac{3}{2} + c_2$$

which was correct, but I can't get the same answer using the method I first mentioned. Here is my work:

$$y'y''=1$$ $$v^2\frac{dv}{dy}=1$$ $$\frac{v^3}{3} = y + c_1$$ $$\frac{dy}{dx} = (y+c_1)^\frac{1}{3}$$ $$x = \frac{3}{2}(y+c_1)^\frac{2}{3} + c_2$$

However this in terms of y is not the same as the answer I got earlier. Could anyone point me in the right direction?

The second question is:

$$y''+y=0$$

I've also been told to solve this using the same method. The only way I've ever solved the before is inspection. Knowing that $\sin(x)$ and $\cos(x)$ fit and putting them into $$y = c_1y_1(x) + c_2y_2(x)$$ which is $$y=c_1\sin(x) + c_2\cos(x)$$.

However trying to solve it the way I've indicated I get:

$$v\frac{dv}{dy}=-y$$ $$vdv = -ydy$$ $$v = \sqrt{c_1-y^2}$$ $$\frac{dy}{dx} = \sqrt{c_1-y^2}$$ $$x = \int{\frac{dy}{\sqrt{c_1-y^2}}}$$ $$x = \arctan(\frac{y}{\sqrt{c_1-y^2}}) + c_2$$

And then I'm just lost

• My textbook was unclear so I referenced this page: mathworld.wolfram.com/… Which said let $v=y'$ – m0meni Feb 25 '15 at 19:36
• I updated my post to show my work. Could you take a look and see where I went wrong? – m0meni Feb 25 '15 at 19:47
• Thank you! This fixes it. However, I have a followup question that I would like to post if you could wait for me to edit my post. – m0meni Feb 25 '15 at 19:53