Question

Find the solution of the given initial value problem. Then plot a graph of the solution. \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=3 t+4, \quad y(0)=y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=y^{\prime \prime \prime}(0)=1\)

Short answer

Based on the steps provided in the solution, summarize your understanding of the process followed to find the unique solution of the given DE. First, the characteristic equation was identified and its roots were found. Next, the complementary function and particular solution were written down. The complementary function and the particular solution were then combined to find the general solution of the DE. After that, the given initial conditions were used to find the unique solution. Finally, the unique solution was plotted using graphing software. Overall, the unique solution was found to be: $$ y(t) = e^{-t} + te^{-t} + \frac{1}{5}\sin t + \frac{2}{5}\cos t $$

Step-by-step solution

Identify the characteristic equation

The given DE is: $$ y^{\mathrm{w}}+2 y^{\prime \prime}+y=\sin t $$ The characteristic equation for this DE is: $$ r^3 + 2r^2 + r = 0 $$

Find the roots of the characteristic equation

First, factor out r from the equation: $$ r(r^2 + 2r + 1) = 0 $$ Now we have a quadratic equation inside the parenthesis: $$ r^2 + 2r + 1 = 0 $$ This is a perfect square trinomial, which can be expressed as: $$ (r + 1)^2 = 0 $$ So, the roots are r=-1 (triple root).

Write the complementary function

Since we have a triple root, the complementary function (CF) will be in the form: $$ CF(y) = C_1e^{-t} + C_2te^{-t} + C_3t^2e^{-t} $$

Write down the particular solution

Since the DE involves a sinusoidal function, we'll consider the particular solution (PS) in the form: $$ PS(y) = A\sin t + B\cos t $$ Now we need to find A and B. Differentiate PS(y) twice with respect to t: $$ PS^{\prime}(y) = A\cos t - B\sin t \\ PS^{\prime \prime}(y) = -A\sin t - B\cos t \\ $$ Now, substitute the PS equation into the given DE. $$ (-A\sin t - B\cos t) + 2(-A\sin t - B\cos t) + (A\sin t + B\cos t)=\sin t \\ $$ Combine the sine and cosine terms: $$ (-A+2B)\sin(t) + (-B - 2A)\cos(t) = \sin t \\ $$ Comparing coefficients: $$ - A + 2B = 1, \\ - B - 2A = 0. $$ Solve the system of linear equations: $$ A = \frac{1}{5} \\ B = \frac{2}{5} $$ Now, the particular solution is: $$ PS(y) = \frac{1}{5}\sin t + \frac{2}{5}\cos t $$

Combine the complementary function and the particular solution

The general solution of the DE given the initial conditions is: $$ y(t) = CF(y) + PS(y) = C_1e^{-t} + C_2te^{-t} + C_3t^2e^{-t} + \frac{1}{5}\sin t + \frac{2}{5}\cos t $$

Use the given initial conditions to find the unique solution

The initial conditions are: $$ y(0)=2, \\ y^{\prime}(0)=0, \\ y^{\prime \prime}(0)=-1, \\ y^{\prime \prime \prime}(0)=1 \\ $$ Applying the initial conditions and solving for the constants: $$ C_1 + \frac{1}{5}\sin(0) + \frac{2}{5}\cos(0) = 2 \\ C_2 - C_1 + \frac{1}{5}\cos(0) - \frac{2}{5}\sin(0) = 0 \\ 2C_2 - 2C_1 + \frac{1}{5}\sin(0) - \frac{2}{5}\cos(0) = -1 \\ -C_2 - 2C_3 - \frac{1}{5}\cos(0) - \frac{2}{5}\sin(0) = 1 \\ $$ Solving this system of linear equations, we find: $$ C_1 = 1, \\ C_2 = 1, \\ C_3 = 0. $$ Now the unique solution is $$ y(t) = e^{-t} + te^{-t} + \frac{1}{5}\sin t + \frac{2}{5}\cos t $$

Plot a graph of the solution

To plot a graph of the solution, we can use graphing software (e.g., Desmos, Wolfram Alpha, etc.) with the equation: $$ y(t) = e^{-t} + te^{-t} + \frac{1}{5}\sin t + \frac{2}{5}\cos t $$ A graph of the solution can be obtained through any online plotting tools by inputting the given function.