I'm stuck on this problem:
Find the solution of the given initial value problem: $ty′+4y=t^2−t+5$, $y(1)=7$, $t>0$
When I multiply both sides by $\mu(t)$ I find that $\mu(t) = e^{4t}$
So:
$$ \frac{d}{dt} (\mu y) = \mu t^2 - \mu t + 5 \mu\\ e^{4t} y= \int t^2 e^{4t}-te^{4t}+5e^{4t} dt $$ This is what I have so far
On
The DE
$$ t y'+4y=t^2-t+5 $$
is linear so it's solution can be composed as
$$ y = y_h+y_p $$
where
$$ ty_h'+4y_h = 0\\ ty_p'+4y_p=t^2-t+5 $$
and
$$ ty_h'+4y_h = 0\Rightarrow \frac{dy_h}{y_h}+4\frac{dt}{t}=0\Rightarrow y_h = \frac{C_1}{t^4} $$
regarding the particular solution taking $y_p = a t^2+b t + c$ we have
$$ t^2(6a-1)+t(5b+1)+4c-5=0\Rightarrow y_p = \frac{t^2}{6}-\frac{t}{5}+\frac 54 $$
hence
$$ y = \frac{C_1}{t^4}+\frac{t^2}{6}-\frac{t}{5}+\frac 54 $$
now considering the initial conditions
$$ y(1) = C_1+\frac 16-\frac 15+\frac 54=7 $$
For $t y′ + 4y = t^2 − t + 5$ then \begin{align} \frac{1}{\mu} \frac{d}{dt} \, ( \mu \, y) &= y' + \frac{\mu'}{\mu} \, y \end{align} which leads to $$ \frac{d}{dt} \ln \mu = \frac{4}{t} = \frac{d}{dt} (4 \ln t) = \frac{d}{dt} \ln(t^4),$$ or $\mu(t) = t^4$. Now, \begin{align} y' + \frac{4}{t} \, y &= t - 1 + \frac{5}{t} \\ \frac{1}{t^4} \, \frac{d}{dt} \, (t^4 \, y) &= t - 1 + \frac{5}{t} \\ \frac{d}{dt} \, (t^4 \, y) &= t^5 - t^4 + 5 \, t^3 \\ y(t) &= t^{-4} \, \int^{t} (u^{5} - u^{4} + 5 \, u^{3}) \, du \\ &= \frac{t^2}{6} - \frac{t}{5} + \frac{5}{4} + \frac{c_{0}}{t^4}. \end{align} By applying $y(1) = 7$ leads to $$y(t) = \frac{10 t^6 - 12 t^5 + 75 t^4 + 347}{60 \, t^4}.$$