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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 + Ce3x is a general solution to the differential equation y 3y = 6. Find the particular solution with the initial condition y(0) = 3.

You have:

y = 2 + Ce3x

Use the initial conditions and get 3 = 2 + Ce30 = 2 + C C = 5

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

y = 2 + 5e3x

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