I have a homogenous differential equation

$a_0 y'' + a_1 y' + a_2 y = 0$

I know that $\lambda_0$ is a double root in characteristic polynomial.

Now I have to show that $y(t) = t e^{\lambda_0 t}$ is a solution to the differential equation.

I cannot just insert the function $y(t)$ in the differential equation because of all the unknown coefficients, but I know that when the second order polynomial has a double root the discriminant is zero and the solutions are

$t = -\frac{b}{2a} = -\frac{a_1}{2 a_0}$.

Can I use this to show that the function is a solution to the diff. eq.?

  • $\begingroup$ Dmoreno: I thought it was best if I added a new question instead of asking multiple questions in the same post. mvw: Fishing for more time? $\endgroup$ – Jamgreen Sep 5 '14 at 9:25

You can simply substitute $y(t)$ into the equation, but as you say, it appears you need some extra information: In this case, the only other information you have is that $\lambda$ is a double root of the characteristic equation $a_0 r^2 + a_1 r + a_2$; in particular, a quadratic polynomial $A r^2 + B r + C$ has a double root iff its discriminant $B^2 - 4 A C$ is zero. Using this will allow you to simplify the expression you produce when substituting the candidate solution $y(t)$.


I don't know why you don't have the coefficients $a_i$, but that solution would pretty much restrict them: $$ \begin{align} y(t) &= t e^{\lambda_0 t} \Rightarrow \\ y'(t) &= e^{\lambda_0 t} + \lambda_0 t e^{\lambda_0 t} \Rightarrow\\ y''(t) &= 2\lambda_0 e^{\lambda_0 t} + \lambda_0^2te^{\lambda_0 t} \end{align} $$ inserting those into the DE gives the equation $$ \begin{align} 0 &= (2\lambda_0 + \lambda_0^2 t)\, a_0 + (1+\lambda_0 t)\, a_1 + t\, a_2 \\ &= \underbrace{2\lambda_0 a_0 + a_1}_0 + (\underbrace{\lambda_0^2 a_0 + \lambda_0 a_1 + a_2}_0) t \end{align} $$ which implies $$ \begin{align} a_1 &= -2\lambda_0\, a_0 \\ a_2 &= \lambda_0^2\, a_0 \end{align} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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