After dealing with first-order equations, we now look at the simplest type of second-order differential equation, with linear coefficients of the form
$$ a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0 \,. $$
To solve this we look at the solutions to the auxiliary equation, given by
$$ ak^2 + bk + c = 0 \,. $$
Based on the solutions of the auxiliary equation, the solution takes the following forms:
If the solutions are $k=\alpha,\beta$ where $\alpha,\beta \in \Re$, with $\alpha \neq \beta$ then the solutions to the original equation are of the form $y=Ae^{\alpha x}+Be^{\beta x}$, where $A,B \in \Re$ are constants of integration to be determined by initial or boundary conditions.
If the solutions are both $k=\alpha$ where $\alpha \in \Re$ then the solutions to the original equation are of the form $y=(A+Bx)e^{\alpha x}$, where $A,B \in \Re$ are constants of integration to be determined by initial or boundary conditions.
If the solutions are $k=\pm \beta i$ where $\beta \in \Re$ then the solutions to the original equation are of the form $y=A\sin(\beta x)+B\cos(\beta x)$, where $A,B \in \Re$ are constants of integration to be determined by initial or boundary conditions.
If the solutions are $k=\alpha\pm\beta i$ where $\alpha,\beta \in \Re$ then the solutions to the original equation are of the form $y=e^{\alpha x}[A\sin(\beta x)+B\cos(\beta x)]$, where $A,B \in \Re$ are constants of integration to be determined by initial or boundary conditions.
We wish to solve
$$ \frac{d^2y}{dx^2}+2\frac{dy}{dx}+y=0 $$
subject to
$$ y(0)=0 \quad \textrm{and} \quad y'(1)=1 \,.$$
The solutions of the auxiliary equation are both $k=−1$, therefore the general solution is
$$ y=(A+Bx)e^{−x} \,,$$
Applying the boundary conditions $y(0)=0,y'(1)=1$ gives $B=e^1,\;A=0$, and the solution is therefore
$$ y(x)=xe^{1−x} \,. $$