Given the general solution to t^2y'' - 4ty' + 4y = 0 is y= c_1t + c_2t^4, how do I solve the problem t^2y'' - 4ty' + 4y = -2t^2 ,   y(1) = 2,    y'(1) =0?

Feb 2, 2015

The answer is: $y = - {t}^{4} + {t}^{2} + 2 t$.

If it is known a solution of the homogenous equation, now a particular solution of the non-homogenous equation has to be find.

Since then in the second member there is $- 2 {t}^{2}$, for the similarity principle the particular solution is a square equation like this:

$\beta \left(t\right) = a {t}^{2} + b t + c$.

Now we calculate:

$\beta ' \left(t\right) = 2 a t + b$, and

$\beta ' ' \left(t\right) = 2 a$.

Now we have to find $a , b , c$.

$\beta \left(t\right)$ has to satisfy the equation, so:

${t}^{2} \left(2 a\right) - 4 t \left(2 a t + b\right) + 4 a {t}^{2} + 4 b t + 4 c = - 2 {t}^{2}$
$2 a {t}^{2} - 8 a {t}^{2} - 4 b t + 4 a {t}^{2} + 4 b t + 4 c = - 2 {t}^{2}$
$- 2 a {t}^{2} + 4 c = - 2 {t}^{2}$

Then, for the principle of the identity between polynomials:

$- 2 a = - 2 \Rightarrow a = 1$, and
$4 c = 0 \Rightarrow c = 0$.
$b$ is not important, so we can assume that is 0.

So:

$\beta \left(t\right) = {t}^{2}$.

The solution of the non-homogeneous is:

$y = {c}_{1} t + {c}_{2} {t}^{4} + {t}^{2}$

and its derivative is:

$y ' = {c}_{1} + 4 {c}_{2} {t}^{3} + 2 t$

Now we have to find ${c}_{1}$ and ${c}_{2}$ using the conditions:

$y \left(1\right) = 2$ and $y ' \left(1\right) = 0$.

$y ' = {c}_{1} + 4 {c}_{2} {t}^{3} + 2 t$.

So:

$2 = {c}_{1} + {c}_{2} + 1 \Rightarrow {c}_{1} + {c}_{2} = 1$ and

$0 = {c}_{1} + 4 {c}_{2} + 2 \Rightarrow {c}_{1} + 4 {c}_{2} = - 2$.

We can substitute ${c}_{1} = 1 - {c}_{2}$ from the first equation to the second:

$1 - {c}_{2} + 4 {c}_{2} = - 2 \Rightarrow 3 {c}_{2} = - 3 \Rightarrow {c}_{2} = - 1$

And:

${c}_{1} = 1 - \left(- 1\right) \Rightarrow {c}_{1} = 2$.

So, finally:

$y = - {t}^{4} + {t}^{2} + 2 t$.