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Second-Order Linear Inhomogeneous Equations

Additional Examples

Solve the initial value problem $$ \begin{align} \frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 20y &= \exp(3x) \\ y(0) &= -6 \\ y'(0) &= 2 \end{align} $$

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. $$ (D^2 - 8D + 20)y = 0 $$ Substep 2: Find the roots. We use the quadratic formula to compute the roots, $ 4\pm 2i$

Substep 3: Write the homogeneous solutions. $$ y_h(x) = c_1\exp(4x)\cos(2x) + c_2\exp(4x)\sin(2x) $$ Step 2: Guess the form of the particular solutions

Since the right-hand side is $\exp(3x),$ we guess $ y_p(x) = A\exp(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 $ 5A\exp(3x) = \exp(3x).$ This gives us the equation $5A=1,$ from which we get $ A=1/5.$ So we get $$ y_p(x)=(1/5)\exp(3x). $$ 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 $$ y(x) = (1/5)\exp(3x) + c_1\exp(4x)\cos(2x) + c_2\exp(4x)\sin(2x) $$ 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) = (3/5)\exp(3x) + 4c_1\exp(4x)\cos(2x) - 2c_1\exp(4x)\sin(2x) + 4c_2\exp(4x)\sin(2x) + 2c_2\exp(4x)\cos(2x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = 1/5 + c_1 &= -6 \\y'(0) = 3/5 + 4c_1 + 2c_2 &= 2 \end{align} $$ We solve these equations to get $ c_1=-31/5$ and $ c_2=131/10.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = (1/5)\exp(3x) - (31/5)\exp(4x)\cos(2x) + (131/10)\exp(4x)\sin(2x). $$ You may reload this page to generate additional examples.


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