Kansas State University

This is a second-order linear constant-coefficient initial value problem.

First, we find the general solution

Step 1: Find the homogeneous solution.

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

Substep 3: Write the homogeneous solutions. $$ y_h(x) = c_1\exp(7x) + c_2\exp(-x) $$ Step 2: Guess the form of the particular solutions

Since the right-hand side is $2\exp(2x),$ we guess $ y_p(x) = A\exp(2x).$

Step 3: Plug our guess into the equation and solve for the undetermined coefficients

Plugging our guess for $ y_p(x)$ into the equation, we obtain $ -15A\exp(2x) = 2\exp(2x).$ This gives us the equation $-15A=2,$ from which we get $ A=-2/15.$ So we get $$ y_p(x)=-(2/15)\exp(2x). $$ Step 4: The general solution is the particular solution plus all the homogeneous solutions.

So from the results of steps 1 and 3, we get the general solution is $$ y(x) = -(2/15)\exp(2x) + c_1\exp(7x) + c_2\exp(-x) $$ Second, we solve for the constants

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) = -(4/15)\exp(2x) + 7c_1\exp(7x) - c_2\exp(-x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = -2/15 + c_1 + c_2 &= -3 \\y'(0) = -4/15 + 7c_1 - c_2 &= 3 \end{align} $$ We solve these equations to get $ c_1=1/20$ and $ c_2=-35/12.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = -(2/15)\exp(2x) + (1/20)\exp(7x) - (35/12)\exp(-x). $$ You may reload this page to generate additional examples.