# Finding homogenous 2nd order differential equation if given general solution

I'm given the general solution to a differential equation: $$y(x)= c_1 \dfrac{\sin(x)}{\sqrt {x}} + c_2 \dfrac{\cos(x)}{\sqrt{x}}$$

This is a uni past paper and I'm absolutely stuck as the homogenous equation in question obviously has non-constant coefficients... Any help would be greatly appreciated.

P.S. Typing this from the phone app, sorry for the lack of LaTeX, if a mod sees this please edit as appropriate.

• @Amzoti It's sin(x)/sqrt(x) and cos(x)/sqrt(x) – Djordje Nov 13 '14 at 22:11

You know that $u(x)=c_1\sin(x)+c_2\cos(x)$ is the general solution of a particular second order ODE, right? So given that $y(x) = u(x) / \sqrt{x}$, try differentiating that twice to see what the ODE for $y$ would look like.

EDIT:

Ok, you say you've got it in a $y''=f(x)$ form, so let's see where we go from there.

We're setting $y=u/\sqrt{x}$ (or equivalently $u=y\sqrt{x}$, and we also know that $u''=-u$. Taking derivatives and doing some rearrangement gives us:

$\begin{eqnarray}u(x) & = & y(x)\times\sqrt{x} \\ & = & y(x)\times x^\frac{1}{2} \\ u'(x) & = & y'(x)\times x^\frac{1}{2} + y(x)\times \frac{1}{2}x^{-\frac{1}{2}} \\ u''(x) & = & y''(x)\times x^\frac{1}{2} + 2 y'(x)\times \frac{1}{2}x^{-\frac{1}{2}} + y(x) \times \frac{1}{2}\times -\frac{1}{2} x^{-\frac{3}{2}} \\ -u(x) & = & y''(x) \times x^\frac{1}{2} + y'(x) \times x^{-\frac{1}{2}} - \frac{1}{4}y(x)\times x^{-\frac{3}{2}} \\ -y(x)\times x^\frac{1}{2} & = & y''(x) \times x^\frac{1}{2} + y'(x) \times x^{-\frac{1}{2}} - \frac{1}{4}y(x)\times x^{-\frac{3}{2}} \end{eqnarray}$

Where in the second-to-last line we've used what we know about $u$, and in the last line we've used the relationship between $u$ and $y$. Finally, we can collect some like terms and get rid of all those fractional powers of $x$, giving us:

$x^2 y''+xy'+\left(x^2 - \frac{1}{4}\right)y = 0$

• I did it and it comes down to a y''=f(x) form - but I need to get a homogenous 2nd order ODE. – Djordje Nov 13 '14 at 23:05
• Figured out what you wanted to say, solved it , thanks! :) – Djordje Nov 13 '14 at 23:59
• Haha, and I finished it off for you anyway! – ConMan Nov 14 '14 at 0:13
• Cheers anyway, it was only a matter of making that leap which I did make with your hint :) – Djordje Nov 14 '14 at 1:37
• No problem. Glad to have been of help :) – ConMan Nov 14 '14 at 2:20

Hint: $u = y/\sqrt{x}$ is the general solution of ...

Function $\tilde y(x)=y\sqrt{x}$ is solution of the differential equation: $$\ddot {\tilde y} +\tilde y=0$$ Thus $$(y \sqrt{x})''+y \sqrt{x}=0$$ Finally we have $$y''\sqrt{x}+y'/2/\sqrt{x}-y/4/x^{3/2}+y \sqrt{x}=0$$

Suppose that two functions $$y_1,y_2$$ are a basis for the solution set of a second-order DE. We can find such an equation by noting that if there is one, $$y_1''+py_1'+qy_1 = 0 = y_2''+py_2'+qy_2 ,$$ which we can treat as the simultaneous equations $$py_1'+qy_1 = -y_1'' \\ py_2'+qy_2 = -y_2''$$ for $$p$$ and $$q$$, which have solution $$p = -\frac{y_1 y_2''-y_1'' y_2}{y_1 y_2'-y_1' y_2} = -\frac{W'}{W} \\ q = \frac{y_1' y_2''-y_1'' y_2'}{y_1 y_2'-y_1' y_2} = \frac{y_1' y_2''-y_1'' y_2'}{W} ,$$ and this is consistent provided that $$W \neq 0$$, which is the case at least when $$y_1$$ and $$y_2$$ are independent and one of them is nonzero. (There is no nice way to write $$y_1' y_2''-y_1'' y_2'$$ in terms of simpler quantities; it is simply $$W(y_1',y_2')$$.)

In this case, $$y_1 = x^{-1/2}\sin{x}$$ and $$y_2 = x^{-1/2}\cos{x}$$, and if we calculate the first couple of derivatives and do the simplifications, we find $$W = - 1/x$$, and then $$p = \frac{1}{x} \qquad q = 1 - \frac{1}{4x^2} .$$