Differential Equations problem Find the general solution of the following equation by substituting $u(t) = y'(t)$
$$ty''+4y'=t^2$$
 A: Hints
We rewrite the system as:
$$u' + \dfrac{4u}{t} = t$$


*

*Use integrating factor to solve this for $u(t)$.

*Substitute and solve $y' = u$


Spoilers - Do Not Peek

 $u(t) = \dfrac{t^6 + 6c}{6t^4}$, $y(t) = \dfrac{t^3}{18} + \dfrac{c}{t^3} + c$

A: As a warm up for solving your equation, plug in 
$u(t) =\mathrm e^{- \int^t P(\tau) \, d\tau}\cdot \left(\int^t \mathrm e^{\int^\tau P(\epsilon) \, d\epsilon}Q(\tau) \, {d\tau} +c \right)$ 
into 
$u'(t) + P(t)\ u=Q(t)$.
A: Given that $$ty''+4y'=t^2$$ and $$u=y'$$ we can rewrite this ODE as $$tu'+4u=t^2$$ which simplifies to $$u'+{4\over t}u=t.$$ We can use the integrating factor $I(x)=e^ {\int P(x)dx}$ to solve this ODE. Let$I(x)=t^4$ and multiply both sides of the equation by $t^4$. This gives us $$t^4u'+4t^3u=t^5$$ which can be rewritten as $${d\over dx}(t^4u)=t^5.$$ Now we must integrate both sides with respect to $t$ $$t^4u =\int t^5dt$$ whch gives us $$t^4u={t^6\over 6} +{C}$$ where $C$ is a constant. Solving for $u$ we get $$u={t^2\over 6}+{C\over t^4}.$$ Since $u=y'$ we see that $$y={t^3\over 18}-{C\over 3t^3}+K$$ where $K$ is a constant. Thus this is the desired solution to the ODE $$ty''+4y'=t^2.$$
