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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.?

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  • $\begingroup$ 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
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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 $y(t)$.

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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 $$ \begin{align} a_1 &= -2\lambda_0\, a_0 \\ a_2 &= \lambda_0^2\, a_0 \end{align} $$

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