Question

A first order linear equation in the form y'+p(x)y=f(x) can be solved by finding an integrating factor mu(x) = exp( int (p(x) dx))
(1) Given the equation y' + 3 y = 2 find mu(x) =
(2) Then find an explicit general solution with arbitrary constant C.
y =
(3) Then solve the initial value problem with y(0)=2
y =

Answer 1

\[y'+3y=2~~\implies~~\mu(x)=e^{\int 3~dx}=e^{3x}\]
\[\begin{align*}e^{3x}y'+3e^{3x}y=2e^{3x}~~\iff~~\frac{d}{dx}\left[e^{3x}y\right]&=2e^{3x}\\
e^{3x}y&=2\int e^{3x}~dx\\
e^{3x}y&=\frac{2}{3}e^{3x}+C\\
y&=\frac{2}{3}+Ce^{-3x}
\end{align*}\]
\[2=\frac{2}{3}+Ce^{-3(0)}~~\iff~~\frac{4}{3}=C~~\implies~~y=\frac{2}{3}+\frac{4}{3}e^{-3x}\]