Differential Equations µ substitution My professor has removed the lecture notes which were our only resource (no textbook) for the class, so I do not have much to go on except for a class I took years ago. He never named it in class, but I remember calling the solution technique µ sub. I can't find any tutorials that I recognize as being this kind of problem or using the equation he provides.
Solve:
$v'=5-0.1v$
with $v(0)=44$
The hint provided is to use the equation:
$y(x)=e^{-µ(x)}\int{e^{-µ(x)}q(x)dx}$
 A: $$v'=5-0.1v$$
For a DE of this kind ( first order linear DE), the integrating factor $\mu (x) $ is given by:
$$v'+p(x)v=q(x) \implies \mu (x) = \exp \int(p(x))dx$$
Here is a step by step solution:
$$v'+0.1v=5$$
Multiply by integrating factor $\mu (x)= \exp \int(0.1)dx=e^{0.1x}$
$$v'(e^{0.1x})+v(0.1e^{0.1x})=5e^{0.1x}$$
Note that:
$$f'g+fg'=(fg)'$$
$$\implies v'(e^{0.1x})+v(0.1e^{0.1x})=(ve^{0.1x})'$$
So that we have:
$$(ve^{0.1x})'=5e^{0.1x}$$
Now integrate both side
$$ve^{0.1x}=5\int e^{0.1x}dx$$
$$ve^{0.1x}=5\frac 1 {0.1} e^{0.1x}+C$$
$$ve^{0.1x}=50 e^{0.1x}+C$$
Finally multiply both sides by $e^{-0.1x}$:
$$v(x)=e^{-0.1x}50 e^{0.1x}+Ce^{-0.1x}$$
$$ \implies v(x)=50+Ce^{-0.1x}$$

For the formula you are given it's the same as what we did:
$$y'+p(x)y=q(x)$$
Multiply both side by $IF (x) =\exp \int p(x) dx$
Your teacher seems to have taken:
$$\mu = \int p(x)dx \implies IF(x)=e^{\mu (x)}$$
$$y'e^{\mu (x)}+y(x)p(x)e^{\mu (x)}=q(x)e^{\mu (x)}$$
Remember that
$$f'g+fg'=(fg)'$$
$$(ye^{\mu (x)})'=q(x)e^{\mu (x)}$$
Integrate both side:
$$ye^{\mu (x)}= \int q(x)e^{\mu (x)}dx$$
Multiply  both side by $e^{-\mu (x)}$
$$y(x)=e^{-\mu (x)} \left ( \int q(x) e^{\mu (x)}dx \right )$$

$$y'+p(x)y=q(x)$$
$$v'+0.1v=5$$
Now apply the given formula but note that $p(x)=0.1$ and $q(x)=5$. Evaluate $\mu (x) = \int p(x)=\int 0.1dx=0.1x$
$$
\begin{align}
y(x)=&e^{-\mu (x)} \left ( \int q(x) e^{\mu (x)}dx \right ) \\
y(x)=&e^{-0.1x} \left ( \int 5 e^{0.1x}dx \right ) \\
y(x)=&e^{-0.1x} \left ( \frac 5 {0.1} e^{0.1x}+C \right ) \\
y(x)=&50+Ce^{-0.1x} \\
\end{align}
$$
