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Second-Order Linear Homogeneous Equations

Additional Examples

Solve the initial value problem $$ \begin{align} \frac{d^2y}{dx^2} + 6\frac{dy}{dx} - 7y &= 0 \\ y(0) &= 3 \\ y'(0) &= 1 \end{align} $$

This is a second-order linear constant-coefficient initial value problem. First we find the general solution.

Step 1: Write the equation in operator form. $$ (D^2 + 6D - 7)y = 0 $$ Step 2: Find the roots. The equation factors into $(D - 1)(D + 7)y = 0.$ So our roots are 1 and -7.

Step 3: Write the general solution.

$$ y(x) = c_1\exp(x) + c_2\exp(-7x) $$ Now that we have the general solution, we plug in our initial values to find the solution to the initial value problem. First we compute, $ y'(x) = c_1\exp(x) - 7c_2\exp(-7x).$ Then we plug in $ x=0 $ to get the following equations. $$ \begin{align} y(0) = c_1 + c_2 &= 3 \\ y'(0) = c_1 - 7c_2 &= 1 \end{align} $$ We solve these equations to get $ c_1= 11/4 $ and $ c_2 = 1/4.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem.

$$ y(x) = (11/4)\exp(x) + (1/4)\exp(-7x) $$ You may reload this page to generate additional examples.


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