# general solution of the differential equation

I have an interesting question.

We have the differential equation $y'+ay=b(x), a \in \mathbb{R}, b(x)$ continuous function in an interval $I$.

We stabilize a $x_0 \in I$.

Then the function $\phi(x)=ce^{-ax}+e^{-ax} \int_{x_0}^x e^{at} b(t) dt \$ for any $c \in \mathbb{R}$ is a solution of the differential equation and conversely each solution of the differential equation is of the above form.

My question:

• $ce^{-ax}$ is a solution of the respective homogeneous differential equation.

• $e^{-ax} \int_{x_0}^x e^{at} b(t) dt$ is a solution of our differential equation.

Why is the sum of the solution of the respective homogeneous differential equation and a solution of our differential equation the general solution of the differential equation?

• The differential operator $D = \frac{d}{dx} + id$ is linear, in the sense that $D(f+g) = Df + Dg$. So, if we apply it first to $c^{-ax}$ we get zero and applying it next to $e{-ax} \int_{x_0}^x e^{at} b(t) dt$ we get $b(x)$, hence applying it to the sum gives $b(x)$. – msteve Feb 14 '15 at 14:36
• See also "Superposition principle" : en.wikipedia.org/wiki/Superposition_principle – JJacquelin Feb 14 '15 at 14:51

## 2 Answers

the particular solution you have $y_P = e^{-ax}\int_{x_0}^x e^{at}b(t) \, dt$ satisfies the differential and $y(x_0) = 0.$ you need the homogenous solution $y_h = ce^{-ax}$ to take care of initial conditions $y(x_0) \neq 0.$

You can substitute $y(x)=e^{-ax}u(x)$ to obtain $$u'(x)=e^{ax}(y'(x)+ay(x))=e^{ax}b(x)$$ This now is a simple integration $$u(x)=C+\int_{x_0}^x e^{at}b(t)\,dt$$ and the integration constant corresponds in the homogeneous solution after reverting to the form of $y(x)$.