Differential equations

Initial value problems for linear equations with constant coefficients.

How to solve it

In a differential equation the unknown is a function $y(x)$, and the equation involves its derivatives: $y'' - 3y' + 2y = 0$. An initial value problem adds conditions, for example $y(0) = 1$ and $y'(0) = 0$, and with them the solution becomes unique.

Step by step

For $y' = ky$ the solution is $y = Ce^{kx}$; find the constant $C$ from $y(0)$.
For $y'' + py' + qy = 0$ write the characteristic equation $r^2 + pr + q = 0$: replace $y''$ by $r^2$, $y'$ by $r$ and $y$ by $1$.
If it has two different real roots $r_1$ and $r_2$, the general solution is $y = C_1 e^{r_1 x} + C_2 e^{r_2 x}$.
Substitute $x = 0$ into $y$ and into $y'$, and find $C_1$ and $C_2$ from the resulting system.

The coefficient of $y'$ becomes the coefficient of $r$.

The coefficient of $y$ becomes the constant term.

Example: $y'' - 3y' + 2y = 0$ gives $r^2 - 3r + 2 = 0$ with roots $1$ and $2$, so the general solution is $y = C_1 e^{x} + C_2 e^{2x}$.

Common mistakes

Getting the sign of a root in the exponent wrong: the root $r = -2$ gives $e^{-2x}$.
Finding the general solution and forgetting the initial conditions.
Differentiating wrongly when using the second condition: $\left(C_2 e^{2x}\right)' = 2C_2 e^{2x}$.