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Second-Order Linear Inhomogeneous Equations

Additional Examples

Solve the initial value problem $$ \begin{align} \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 13y &= -3\exp(x) \\ y(0) &= 3 \\ y'(0) &= 2 \end{align} $$

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. $$ (D^2 + 4D + 13)y = 0 $$ Substep 2: Find the roots. We use the quadratic formula to compute the roots, $ -2\pm 3i$

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

Since the right-hand side is $-3\exp(x),$ we guess $ y_p(x) = A\exp(x).$

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 $ 18A\exp(x) = -3\exp(x).$ This gives us the equation $18A=-3,$ from which we get $ A=-1/6.$ So we get $$ y_p(x)=-(1/6)\exp(x). $$ 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 $$ y(x) = -(1/6)\exp(x) + c_1\exp(-2x)\cos(3x) + c_2\exp(-2x)\sin(3x) $$ 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) = -(1/6)\exp(x) - 2c_1\exp(-2x)\cos(3x) - 3c_1\exp(-2x)\sin(3x) - 2c_2\exp(-2x)\sin(3x) + 3c_2\exp(-2x)\cos(3x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = -1/6 + c_1 &= 3 \\y'(0) = -1/6 - 2c_1 + 3c_2 &= 2 \end{align} $$ We solve these equations to get $ c_1=19/6$ and $ c_2=17/6.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = -(1/6)\exp(x) + (19/6)\exp(-2x)\cos(3x) + (17/6)\exp(-2x)\sin(3x). $$ You may reload this page to generate additional examples.


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