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:


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

  • $\begingroup$ My textbook was unclear so I referenced this page: mathworld.wolfram.com/… Which said let $v=y'$ $\endgroup$ – m0meni Feb 25 '15 at 19:36
  • $\begingroup$ I updated my post to show my work. Could you take a look and see where I went wrong? $\endgroup$ – m0meni Feb 25 '15 at 19:47
  • $\begingroup$ 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. $\endgroup$ – m0meni Feb 25 '15 at 19:53

First problem, was the missing factor 3 in the 2nd last equation.

The second problem is then \begin{align} \tan(x+c) &= \frac{y}{\sqrt{d-y^2}} \Rightarrow \\ (\tan(x+c))^2 &= \frac{y^2}{d-y^2} \iff \\ (\tan(x+c))^2 (d-y^2) &= y^2 \iff \\ d (\tan(x+c))^2 &= (1+(\tan(x+c))^2) y^2 \end{align} so \begin{align} y &= \pm \frac{\sqrt{d} \tan(x+c)}{\sqrt{1+(\tan(x+c))^2}} \\ &= \pm \frac{\sqrt{d} \tan(x+c)}{\sqrt{\frac{1}{\cos(x+c))^2}}} \\ &= \pm \sqrt{d} \sin(x+c) \\ &= \pm \sqrt{d} \left[\sin(c) \cos(x) + \cos(c) \sin(x) \right] \end{align}

  • $\begingroup$ Thank you very much! That makes a lot of sense. $\endgroup$ – m0meni Feb 25 '15 at 20:57

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