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. $$ (D^2 - 4D + 3)y = 0 $$ Substep 2: Find the roots. The equation factors into $ (D - 3)(D - 1)y = 0.$ So our roots are 3 and 1.

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

Since the right-hand side is $2\cos(3x),$ we guess $ y_p(x) = A\cos(3x) + B\sin(3x).$

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 $ (-6A - 12B)\cos(3x) + (12A - 6B)\sin(3x) = 2\cos(3x).$ This gives us the equations $-6A - 12B=2$ and $12A - 6B=0.$ From these equations we get $ A=-1/15$ and $B=-2/15.$ So we get $$ y_p(x)=-(1/15)\cos(3x) - (2/15)\sin(3x). $$ 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 $$ y(x) = -(1/15)\cos(3x) - (2/15)\sin(3x) + c_1\exp(3x) + c_2\exp(x) $$ 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) = -(2/5)\cos(3x) + (1/5)\sin(3x) + 3c_1\exp(3x) + c_2\exp(x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = -1/15 + c_1 + c_2 &= -9 \\y'(0) = -2/5 + 3c_1 + c_2 &= -6 \end{align} $$ We solve these equations to get $ c_1=5/3$ and $ c_2=-53/5.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = -(1/15)\cos(3x) - (2/15)\sin(3x) + (5/3)exp(3x) - (53/5)\exp(x). $$ You may reload this page to generate additional examples.