# How do you solve the initial value problem of t^2y'' - 4ty' + 4y = -2t^2 given y(1) = 2 and y'(1) =0?

Feb 19, 2015

Answer : $y : t \setminus \mapsto 2 t - {t}^{4} + {t}^{2}$

1) First, solve ${t}^{2} y ' ' - 4 t y ' + 4 y = 0$.

• Consider $y \left(t\right) = {t}^{k}$ where $k$ is real. It's a solution if and only if
${t}^{2} k \left(k - 1\right) {t}^{k - 2} - 4 t k {t}^{k - 1} + 4 {t}^{k} = 0$
or if and only if ${k}^{2} - 5 k + 4 = 0$. The solutions are 1 et 4.

• Therefore, all the solutions of ${t}^{2} y ' ' - 4 t y ' + 4 y = 0$ are

$t \setminus \mapsto \setminus \lambda t + \setminus \mu {t}^{4}$

where $\lambda$ and $\mu$ are arbitrary real constants.

2) Second, find a particular solution of ${t}^{2} y ' ' - 4 t y ' + 4 y = - 2 {t}^{2}$.

• Try a function like $y \left(t\right) = m {t}^{k}$. You find easily that $m = 1$ and $k = 2$ are ok.

• Therefore all the solutions of ${t}^{2} y ' ' - 4 t y ' + 4 y = - 2 {t}^{2}$ are

$y : t \setminus \mapsto \setminus \lambda t + \setminus \mu {t}^{4} + {t}^{2}$

where $\lambda$ and $\mu$ are arbitrary real constants.

3) Finally, you want to have $y \left(1\right) = 2$ and $y ' \left(1\right) = 0$.

• Because $y \left(1\right) = \setminus \lambda + \mu + 1$, you have lambda + mu = & 1

• Because $y ' \left(t\right) = \lambda + 4 \mu {t}^{3} + 2 t$, you have $\lambda + 4 \mu + 2 = 0$.

• You solve the system above. You get $\mu = - 1$ and $\lambda = 2$.

Conclusion The unique solution is $t \setminus \mapsto 2 {t}^{2} - {t}^{4} + {t}^{2}$.