Linear differential equation: $$ a(t)y'(t)+b(t)y(t)+c(t)=0 $$ Any equation you need to write down in this form.

Integrating factor is: $$ \mu(t) = \exp\left( \int^{t} \frac{b(s)}{a(s)}ds\right) = e^{\int^{t} \frac{b(s)}{a(s)}ds} $$

Potential function is: $$ \psi(y,t)=y(t) \mu(t) + \int^{t} \frac{c(s)\mu(s)}{a(s)}ds $$ General solution: $$ y_{g}(t) = -\frac{1}{\mu(t)}\left(\int^{t} \frac{c(s)\mu(s)}{a(s)}ds \right) $$

Example: Consider $y'=2y+1$. Find


1) Integrating factor $\mu$ with inital condition $\mu(0)=1$

2) Potential function $\psi$ without arbitrary constant $C$,

3) General solution $y_{g}$

4) Solution to inital value problem $y(0)=2$


Solution:

Your equation is $y'-2y-1=0$ therefore $a(t)=1, b(t)=-2, c(t)=-1$. So integrating factor is $$ \mu(t) = \exp\left(\int^{t} \frac{b}{a} \right) = \exp\left(\int^{t} -2 \right) =\exp(2t+C)=e^{2t+C} $$ Since $\mu(0)=1$ therefore we get $C=0$ and hence $\mu(t) = e^{2t}$.

Potential function is $$ \psi(y,t) = y \mu + \int^{t} \frac{c \mu }{a} = y e^{2t}+ \int^{t} -1 \cdot e^{2s}ds = ye^{2t}-\frac{e^{2t}}{2}+C $$ Since they are asking us don't include arbitrary $C$ therefore $\psi(y,t)=ye^{2t}-\frac{e^{2t}}{2}$.

General solution is $$ y_{g}(t) =-\frac{1}{\mu(t)}\left( \int^{t} \frac{c \mu }{a}\right)=-\frac{1}{e^{2t}}\left( \int^{t}-1\cdot e^{2s}ds\right)= -\frac{1}{e^{2t}}\left(-\frac{e^{2t}}{2}+C\right) $$ So your general solution is $y_{g}(t)=-\frac{1}{e^{2t}}\left(-\frac{e^{2t}}{2}+C\right)$.
REMARK: Sometimes webwork wants you to change the sign of the constant $C$ in the general solution, for example it wants you to write the general solution in this way $$ y_{g}(t)=-\frac{1}{e^{2t}}\left(-\frac{e^{2t}}{2}-C\right) $$


Solution to the initial value problem $y(0)=2$ means that you need to choose $C$ so that $y_{g}(0)=2$. On the one hand $y_{g}(0)=-1(-\frac{1}{2}-C)$ on the other hand $y_{g}(0)=2$. Therefore $-1(-\frac{1}{2}-C)=2$ if and only if $C=2-\frac{1}{2}=\frac{3}{2}$ Therefore solution to the inital value problem $y(0)=2$ is the following $$ y(t)=-\frac{1}{e^{2t}}\left(-\frac{e^{2t}}{2}-\frac{3}{2}\right) $$





Example2: Find the differential equation of the form $y'=f(y)$ such that one of its solution is $y=3e^{2t}+5$.

Solution:

Equality $y'=f(y)$ means that $6e^{2t}=f(y)$. But equality $y=3e^{2t}+5$ implies that $6e^{2t}=2(y-5)$ therefore equality $6e^{2t}=f(y)$ becomes: $2(y-5) = f(y)$ and this is the answer. Thing is that everyhting you should express in terms of $y$.