Solutions of Differential Equations

A solution of a differential equation is an expression of the dependent variable that satisfies the relation established in the differential equation. For example, the solution of $y'+y-2=4x$ will be some equation y = f(x) such that y and its first derivative, y', satisfy the relation $y'+y-2=4x$. The general solution of a differential equation will have one or more arbitrary constants, depending on the order of the original differential equation (the solution of a first order diff. eq. will have one arbitrary constant, a second order one will have two, etc.).

Examples:

• $y'=2x$. Through simple integration, we can calculate the general solution of this equation to be $y=x^{2}+C$, where C is an arbitrary constant.
• $y'-y=0$. The G.S. is $y=Ce^{x}$. $y'=Ce^{x}$, so $y'-y=0\to Ce^{x}-Ce^{x}=0$, so this solution satisfies the relationship for all arbitrary constants C.
• $y''+y'-2y=0$. The G.S. is $y=Ce^{x}+De^{-2x}$. $y'=Ce^{x}-2De^{-2x}$ and $y''=Ce^{x}+4De^{-2x}$, so $0=y''+y'-2y$ becomes $0=Ce^{x}+4De^{-2x}+Ce^{x}-2De^{-2x}-2(Ce^{x}+De^{-2x})=Ce^{x}+Ce^{x}-2Ce^{x}+4De^{-2x}-2De^{-2x}-2De^{-2x})=0$.

The particular solution of a differential equation can be solved if we have enough points to solve for the arbitrary constants.

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 $y=e^{x}+e^{-2x}$.

Resources

• Differential Equations, University of Glascow
• General and Particular Solutions, Simply Math