Find the general solution to the ODE:$$t^2y''-ty'+y=0~~~~~~~~(*)$$given that $y_1=t~~$is a solution.

My intuitive solution:

Let $x=ln(t)$, then $y'=\frac{dy}{dx}\frac{1}{t}$ and $y''=\frac{d^2y}{dx^2}\frac{1}{t^2}-\frac{dy}{dx}\frac{1}{t^2}$

Plug into $(*)$ we get: $$t^2(\frac{d^2y}{dx^2}\frac{1}{t^2}-\frac{dy}{dx}\frac{1}{t^2})-t(\frac{dy}{dx}\frac{1}{t})+y=0$$$$\rightarrow~~\frac{d^2y}{dx^2}-2\frac{dy}{dx}+y=0$$ So the characteristic equation for this is $r^2-2r+1=0~\rightarrow~(r-1)^2=0~\rightarrow~r=1$.

Then $y_1(x)=y_2(x)=e^x~\rightarrow~y_1(t)=y_2(t)=e^{ln(t)}=t.~$But now the Wronskian is certainly zero. So I tried to multiply a $t$ such that $y_2(t)=t^2.~$But $y'_2=2t~~$and$~~ y''_2=2.~$Substituting these into $(*)$ I found that this doesn't solve the equation.

Could someone please give me a hint on how to find the general solution?

  • $\begingroup$ Since $r=1$ is a double-root try $y = Cxe^x$ for the second solution (the first solution is as you have found $y=e^x$). $\endgroup$ – Winther Mar 12 '15 at 0:26
  • $\begingroup$ Oh...OK, got it, I shouldn't have multiplied the factor to $y_2(t)$ but $y_2(x)$... Thanks! $\endgroup$ – Ruihong Yuan Mar 12 '15 at 0:47

Based on your analysis, as the solutions of the characteristic equation is $r=1$(double root), the two basic solutions for this ODE are $y_1(x)=e^x\rightarrow y_1(t)=t$ and $y_2(x)=xe^x\rightarrow y_2(t)=t\ln|t|$. Substituting these solutions into $(*)$ you will find they are the solutions. And their combinations are also the solutions, i.e., $$y(t)=C_1t+C_2t\ln|t|$$ where $C_1$ and $C_2$ are constants.

Hope this can help you.



Consider the method of the reduction of order.

You know one of the solutions to the differential equation, which is: $y_1 = t$

Now, call the second solution $y_2$ where $y_2 = vy_1 = vt$, where $v$ is a function of $t$.

Taking the derivative, we get:

$$y_2 = vt$$

$$y_2' = v't + v$$

$$y_2'' = v''t + v' + v' = v''t + 2v'$$

Substituting this into our equation:

$$t^2(v''t + 2v') -t(v't + v) + vt = 0$$

$$t^3v'' + 2t^2v' - t^2v' - vt + vt =0,\ t^3v'' + 2t^2v' - t^2v' = 0$$

Note: Upon using this method the $y$ term has dropped out. This method ensures that, and now we can find our second solution:

$$t^3v'' + t^2v' = 0$$

Now, we let $w = v'$, $w' = v''$

It follows that:

$$t^3w' + t^2w = 0$$

$$w' = -\frac{w}{t}$$

$$\frac{dw}{w} = -\frac{dt}{t}$$

$$\ln w = -\ln t$$

$$w = -t,\ v' = -t,\ v = -\frac{t^2}{2}$$

Now, our second solution is $$y_2 = vt = -\frac{t^3}{2}$$

Now, our general solution is formed:

$$y(t) = c_1t + c_2\frac{t^3}{2}$$

  • $\begingroup$ From $\ln w=-\ln t$ it follows $w=1/t$, you went a little fast there. Then $v=\ln|t|$ in accordance with the other answers. $\endgroup$ – Lutz Lehmann Mar 12 '15 at 3:06
  • $\begingroup$ @LutzL ty for catching my mistake $\endgroup$ – Varun Iyer Mar 12 '15 at 23:23

this is an example of eulers equation. try $y = t^k$ sub it in the equation. find $$k(k-1) - k + 1= 0\to k = 1, 1. $$ the two solutions are $$t, t \ln t $$ and the general solution is $$y = At + B t\ln t. $$

  • $\begingroup$ You meant the type is "Euler equation". $\endgroup$ – Lutz Lehmann Mar 12 '15 at 3:07

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