# How to Solve First Order Differential Equations with Initial Conditions

Often you want to find a function that is a solution to a given differential equation which at the same time either passes through a specific point, or has a specific value, for a given $x$-value. This extra requirement is called an initial condition. It enables you to determine the constant of the solution.

When the constant is determined, it is called a particular solution. You find the particular solution by inserting the values of the initial conditions, and of $x$ in the function, into the equation.

Example 1

The function $y=-2+C{e}^{3x}$ is a general solution to the differential equation ${y}^{\prime }-3y=6$. Find the particular solution with the initial condition $y\left(0\right)=3$.

You have:

 $y=-2+C{e}^{3x}$

Use the initial conditions and get $\begin{array}{cc}3=-2+C{e}^{3\cdot 0}=-2+C& \\ C=5& \end{array}$

Insert the value for $C$ back into the general equation and find the particular solution

 $y=-2+5{e}^{3x}$