Initial Value Problem

With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point $(x,y(x))$, for a second order equation we need 2 points (typically $(x_{1},y(x_{1}))$ and either $(x_{2},y(x_{2}))$ or $(x_{2},y'(x_{2}))$), and so on.

Examples:

• $y'=2x$, $y(2)=0$. With this point and the general solution $y=x^{2}+C$, we can calculate the constant C to be -4. Thus the particular solution is $y=x^{2}-4$.
• $y'-y=0$, $y(0)=3$. $3=Ce^{0}\implies C=3$, so the particular solution is $y=3e^{x}$.
• $y''+y'-2y=0$, $y(0)=2$, $y'(0)=-1$. So, $2=Ce^{0}+De^{0}=C+D$ and $-1=Ce^{0}-2De^{0}=C-2D$. Thus C = 1 and D = 1, and the particular solution is $y=e^{x}+e^{-2x}$.