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Second-Order Linear Inhomogeneous Equations

Additional Examples

Solve the initial value problem $$ \begin{align} \frac{d^2y}{dx^2} - \frac{dy}{dx} &= 6\sin(x) \\ y(0) &= 9 \\ y'(0) &= 4 \end{align} $$

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

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

Since the right-hand side is $6\sin(x),$ we guess $ y_p(x) = A\cos(x) + B\sin(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 $ (-A - B)\cos(x) + (A - B)\sin(x) = 6\sin(x).$ This gives us the equations $-A - B=0$ and $A - B=6.$ From these equations we get $A=3$ and $B=-3.$ So we get $$ y_p(x)=3\cos(x) - 3\sin(x). $$ 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 $$ y(x) = 3\cos(x) - 3\sin(x) + c_1 + c_2\exp(x) $$ 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) = -3\cos(x) - 3\sin(x) + c_2\exp(x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = 3 + c_1 + c_2 &= 9 \\y'(0) = -3 + c_2 &= 4 \end{align} $$ We solve these equations to get $ c_1=-1$ and $ c_2=7.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = 3\cos(x) - 3\sin(x) - 1 + 7\exp(x). $$ You may reload this page to generate additional examples.


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