Solving a differential equation?

Solution 1:

Given a non-trivial interval $I$ and $a,b$ continuous functions (therefore integrable) defined on $I$, consider the differential equation $y′+ay=b$. Let $A$ be an antiderivative of $a$. There exists $K\in \mathbb R$ such that for all $t\in I$ the following holds: \begin{align} y'(t)+a(t)y(t)=b(t)&\iff e^{A(t)}(y'(t)+a(t)y(t))=e^{A(t)}b(t)\\ &\iff \dfrac{d}{dt}\left(t\mapsto e^{A(t)}y(t)\right)(t)=e^{A(t)}b(t)\\ &\iff e^{A(t)}y(t)=\int e^{A(t)}b(t)\,\mathrm dt + K\\ &\iff y(t)=e^{-A(t)}\int e^{A(t)}b(t)\,\mathrm dt+Ke^{-A(t)},\end{align} where $\displaystyle \int e^{A(t)}b(t)\mathrm dt$ denotes an antiderivative of $t\mapsto e^{A(t)}b(t)$ on $I$.

As mentioned the solution $v(t)=\dfrac c a$ for all $t$ is given in the general formula.

Solution 2:

"Why isn't this a valid solution to the equation?": it absolutely is, setting $K_1=\frac{c}{a}$ and $K_2=0$ !

You just found a particular case of the general solution.