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

You have:

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

Use the initial conditions and get

\begin{matrix} 3 = - 2 + C e^{3 \cdot 0} = - 2 + C \\ C = 5 \end{matrix}

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

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

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