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>Second Order>How to Solve Second Order Differential Equations with Initial Conditions

How to Solve Second Order Differential Equations with Initial Conditions

Just like with the first order differential equations, you often want to find a function which is the solution to a given differential equation with certain initial conditions. However, for second order differential equations, you need two initial conditions instead of one.

You’ll often be given a point—that is, a functional value at a given $x$ , and a value for the derivative at another point. With these two conditions, you can find the constants of the general solution and get a particular solution. You determine the particular solution by inserting the values for the initial conditions and solve the resulting system of equations with two unknowns.

Example 1

Solve the differential equation $y^{″} + y^{'} - y = 0$ with the initial conditions $y (0) = 2$ and $y^{'} (0) = 1$

This differential equation has the solution

y (x) = C_{1} e^{x} + C_{2} e^{- 2 x}

The derivative of () is

y^{'} (x) = C_{1} e^{x} - 2 C_{2} e^{- 2 x}

To find $C_{1}$ and $C_{2}$ , you make two equations with two unknowns from the initial conditions and solve the system of equations:

\begin{array}{l} 2 & = y (0) \\ = C_{1} e^{0} + C_{2} e^{0} \\ = C_{1} + C_{2} \\ C_{1} & = 2 - C_{2} \\ C_{1} & = 2 - \frac{1}{3} \\ = \frac{5}{3} \\ 1 & = y^{'} (0) \\ = C_{1} e^{0} - 2 C_{2} e^{0} \\ = C_{1} - 2 C_{2} \\ 1 & = (2 - C_{2}) - 2 C_{2} \\ = 2 - 3 C_{2} \\ C_{2} & = \frac{1}{3} \end{array}

\begin{array}{l} 2 & = y (0) & 1 & = y^{'} (0) \\ = C_{1} e^{0} + C_{2} e^{0} & = C_{1} e^{0} - 2 C_{2} e^{0} \\ = C_{1} + C_{2} & = C_{1} - 2 C_{2} \\ C_{1} & = 2 - C_{2} \\ 1 & = (2 - C_{2}) - 2 C_{2} \\ = 2 - 3 C_{2} \\ C_{2} & = \frac{1}{3} \\ C_{1} & = 2 - \frac{1}{3} \\ = \frac{5}{3} \end{array}

Insert the values for

C_{1}

and

C_{2}

into the general solution () to find the particular solution:

y (x) = \frac{5}{3} e^{x} + \frac{1}{3} e^{- 2 x}

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