Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Solve $t^2x''-4tx'+6x=0$ knowing that $x_1(t)=t^2+t^3$ is a particular solution

So I assumed the general solution will be in form of $x(t)=C_1 x_1(t)+C_2 x_2(t)$ and $x_2 = v(t)x_1(t)$

So now I have to find $x_2$

$$x_2=v(t^2+t^3)$$ $$x_2'=v'(t^2+t^3)+v(2t+3t^2)$$ $$x_2''=v''(t^2+t^3)+2v'(2t+3t^2)+v(2+6t)$$

And if I try to plug it into my equation I get something very complicated, as almost nothing reduces. Is it the right way to do it?

EDIT: ill try to plug it in here and see what happens since I probably made a mistake

$$t^2(v''(t^2+t^3)+2v'(2t+3t^2)+v(2+6t))-4t(v'(t^2+t^3)+v(2t+3t^2))+6(v(t^2+t^3))=0$$ dividing both sides by $t^2$ $$v''(t^2+t^3)+2v'(2t+3t^2)+v(2+6t)-4(v'(t+t^2)+v(2+3t))+6(v(1+t))=0$$ $$v''(t^2+t^3)+v'(2(2t+3t^2)-4(t+t^2))+v((2+6t)-4(2+3t)+6(1+t))=0$$ $$v''(t^2+t^3)+v'(4t+6t^2-4t-4t^2)+v(2+6t-8-12t+6+6t)=0$$ $$v''(t^2+t^3)+v'(2t^2)=0$$

Yeah I made a mistake, thanks for pointing that out.

share|cite|improve this question
Try using the Wronskian instead? – Alec Aug 29 '14 at 13:30
Sorry but I have no idea how to do that. – Lugi Aug 29 '14 at 13:38
Most of the $v$ terms should have reduced to $0$ and some of the $v'$ terms as well. Can you show what you got when you tried this? – abiessu Aug 29 '14 at 13:53
If my mental arithmetic is functioning correctly, I get $$v''(t^2+t^2)+2v'=0$$ – abiessu Aug 29 '14 at 13:56

You're close. If you think about it, your particular solution is actually two separate solutions.

$$x_1(t) = t^2$$


$$x_2(t) = t^3$$

These two form a basis for all solutions. This type of ODE can be solved with the ansatz that $x(t) = t^p$ for some power $p$. (Because the derivatives are accompanied by powers of $t$.) So with this guess, you get

$p(p-1) - 4p + 6 =0$

This means $p$ is two or three.

share|cite|improve this answer

This is an "Euler equation". Doing the transformation $$ x(t)=z(\log t), $$ you obtain for $z$ the equation $$ z''-5z'+6z=0, $$ and hence $$ z(t)=c_1\mathrm{e}^{2t}+c_2\mathrm{e}^{3t}, $$ and hence $$ x(t)=c_1 t^2+c_2t^3, $$ which is the general solution of your equation.

share|cite|improve this answer

In standard form, the equation is $x'' - \frac{4}{t}x' + \frac{6}{t^2}x = 0$. The Wronskian is then $W = c_1e^{-\int -\frac{4}{t}dt} = c_1t^4$. Then, since the Wronskian is $W = x_1x_2'-x_1'x_2$, we get a first order ODE. $$ c_1t^4 = (t^2+t^3)x_2'-(2t+3t^2)x_2 \implies c_1t^3 = t(1+t)x_2'-(2+3t)x_2 $$ Again using the method of the integrating factor on that expression yields $$ x_2 = c_2t^2(t+1)-c_1t^2 $$ Picking almost any $c_1$ or $c_2$ should give you a linearly independent solution. For example, with $c_2=0$ and $c_1=-1$, you get $x_2=t^2$.

Edit: here's a link that talks about the Wronskian in second order ODEs.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.