Initial conditions · Differential Equations

Intuition

With the general solution in hand, a value and a slope at the starting point give two linear equations for the two constants. Solving them is a small piece of linear algebra, and it always works — the next lesson explains why the system is never singular. What does not always work is asking for values at two different points: that is a boundary value problem, and it can have one solution, none, or infinitely many.

A shell fired from a gun: where the gun stands, with the angle and speed of the shot, determines the whole flight. Asking instead for the shell to land on a particular spot fixes a condition at the far end, and depending on the target there may be two firing angles, one, or none.

The solution of $y''-3y'+2y=0$ with $y(0)=1$ and $y'(0)=0$ is $y=2e^{x}-e^{2x}$ . The value fixes the point and the slope fixes the tangent there; of the whole two-constant family, exactly one curve does both.

Two conditions, two equations

For $y=C_{1}y_{1}+C_{2}y_{2}$ with and , the constants solve the linear system below. When and are independent solutions its determinant, the Wronskian of the next lesson, is not zero, so the system has exactly one solution.

\begin{pmatrix}y_{1}(x_{0})&y_{2}(x_{0})\\y_{1}'(x_{0})&y_{2}'(x_{0})\end{pmatrix}\begin{pmatrix}C_{1}\\C_{2}\end{pmatrix}=\begin{pmatrix}y_{0}\\v_{0}\end{pmatrix}

Doing it right

$y''-3y'+2y=0$ , , : and , so , and .

Applications

Practice

Differentiate, Then Substitute

Write the general solution and its derivative, put the starting point into both, and solve the two equations.

C_{1}+C_{2}=y_{0},\qquad r_{1}C_{1}+r_{2}C_{2}=v_{0}

Try it

Solve $y''-y=0$ with $y(0)=3$ and $y'(0)=1$ .