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. $$ (D^2 + 13D + 30)y = 0 $$ Substep 2: Find the roots. The equation factors into $ (D + 3)(D + 10)y = 0.$ So our roots are -3 and -10.

Substep 3: Write the homogeneous solutions. $$ y_h(x) = c_1\exp(-3x) + c_2\exp(-10x) $$ Step 2: Guess the form of the particular solutions

Since the right-hand side is $-2\exp(7x),$ we guess $ y_p(x) = A\exp(7x).$

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 $ 170A\exp(7x) = -2\exp(7x).$ This gives us the equation $170A=-2,$ from which we get $ A=-1/85.$ So we get $$ y_p(x)=-(1/85)\exp(7x). $$ 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 $$ y(x) = -(1/85)\exp(7x) + c_1\exp(-3x) + c_2\exp(-10x) $$ 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) = -(7/85)\exp(7x) - 3c_1\exp(-3x) - 10c_2\exp(-10x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = -1/85 + c_1 + c_2 &= 5 \\y'(0) = -7/85 - 3c_1 - 10c_2 &= -4 \end{align} $$ We solve these equations to get $ c_1=33/5$ and $ c_2=-27/17.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = -(1/85)\exp(7x) + (33/5)\exp(-3x) - (27/17)\exp(-10x). $$ You may reload this page to generate additional examples.