Activities for Second Order ODEs

Section 3.5 Activities for Second Order ODEs

Subsection 3.5.1 Homogeneous Equations

Activity 3.5.1.

Consider this SOCCLDE.

\begin{align*} \amp y^{\prime \prime} + 8 y^\prime + 4y = 0 \amp \amp y(0) = 10 \amp \amp y^\prime(0) = -4 \end{align*}

Solve the equation.
You should have found real roots to the quadratic equation above, giving exponential soltuions. Now take the coefficient of \(y^\prime\) term and make it smaller. See what happens to the solutions.
What coefficient of \(y^\prime\) leads to critically damped solutions?
How small must the coefficient of \(y^\prime\) be to lead to oscillations?
If this is a harmonic system, what are you changing? If it is a circuit?

Solution.

First I write and solve the characteristic equation.
\begin{equation*} r^2 + 8r + 4 = 0 \implies r = -4 \pm \sqrt{12} \end{equation*}
The root are real and distinct. Therefore, the homogeneous solutions are exponential.
\begin{equation*} y = A e^{(-4+\sqrt{12})t} + Be^{(-4-\sqrt{12})t} \end{equation*}
I also need the derivatives of the homogeneous solution.
\begin{equation*} y^\prime = (-4+\sqrt{12}) A e^{(-4+\sqrt{12})t} + (-4-\sqrt{12}) B e^{(-4-\sqrt{12})t} \end{equation*}
Then I use the initial conditions.
\begin{gather*} y(0) = A + B = 10 \\ y^\prime(0) = (-4+\sqrt{12}) A + (-4-\sqrt{12}) B = -4 \end{gather*}
I can do the algebra to solve the system. I use the first equation to replace \(A\) with \(10-B\text{.}\) Then I manipulate to solve for \(B\) and finally use the first equation again to calculate \(A\text{.}\)
\begin{align*} -4 \amp = (-4+\sqrt{12}) (10-B) + (-4-\sqrt{12}) B \\ -4 \amp = (-40+10\sqrt{12}) + (4-\sqrt{12})B + (-4-\sqrt{12}) B \\ 36 - 10 \sqrt{12} \amp = -2\sqrt{12})B \\ \frac{36 - 10 \sqrt{12}}{-2\sqrt{12}} \amp = B \\ \frac{5 \sqrt{12}-18}{\sqrt{12}} \amp = B \\ A \amp = 10 - B = \frac{18 + 5 \sqrt{12}}{\sqrt{12}} \end{align*}
Therefore, the specific solution is
\begin{equation*} y = \frac{18 + 5\sqrt{12}}{\sqrt{12}} e^{(-4+\sqrt{12})t} + \frac{5 \sqrt{12} - 18}{\sqrt{12}} e^{(-4-\sqrt{12})t} \end{equation*}
When I make the coefficient of \(y^\prime\) smaller, it decreases the magnitude of the positive term \(\sqrt{b^2 - 4ac}\) in the quadratic equation. Therefore, the roots become closer and closer together. For small changes in this coefficient, the roots remain real and the solutions remain exponentia.
Critically damped happens when the solution to the characteristic equation is a repeated root. This happens exactly when \(b^2-4ac = 0\text{.}\) In this equation, \(4ac = 16\text{,}\) so the coefficient of \(y^\prime\) must shink to \(4\) to get this behaviour.
To get oscillations, you need to keep decreasing the coefficient bast the critically damped point. With the logical of the previous point, oscillations will start when the coefficient is decreased below \(4\text{.}\)
In a harmonic system, you are changing the friction term. By lowering the friction, you allow for more movement and eventually for oscillations. in a circuit, you are lowering the resistance, which is similar to friction in that it removes energy from the system.

Activity 3.5.2.

Consider this SOCCLDE.

\begin{align*} \amp y^{\prime \prime} + 4 y^\prime + 20y = 0 \amp \amp y(0) = 2 \amp \amp y^\prime(0) = 6 \end{align*}

Solve the equation.
You should have found complex roots to the quadratic equation above, giving sinusoidal soltuions. Now take the coefficient of \(y\) term and make it smaller. See what happens to the solutions.
What coefficient of \(y\) leads to critically damped solutions?
How small must the coefficient of \(y^\prime\) be to lead to exponential decay?
If this is a harmonic system, what are you changing? If it is a circuit?

Solution.

First I write and solve the characteristic equation.
\begin{equation*} r^2 + 4r + 20 = 0 \implies r = -2 \pm \sqrt{-16} \end{equation*}
The root are complex with a negative real part. \(\alpha = -2\) and \(\beta = 4\text{.}\) Therefore, the homogeneous solutions are sinusoidal with exponentially decay in amplitude.
\begin{equation*} y = A e^{-2t} \cos 4t + B e^{-2t} \sin 4t \end{equation*}
I also need the derivatives of the homogeneous solution. (I’ve grouped like terms after taking the derivative).
\begin{equation*} y^\prime = (-2A + 4B) e^{-2t} \cos 4t + (-4A -2B) e^{-2t} \sin 4t \end{equation*}
Then I use the initial conditions.
\begin{align*} y(0) \amp = A = 2 \\ y^\prime(0) \amp = (-2A + 4B) = 6 \\ -4 + 4B \amp = 6 \implies B = \frac{10}{4} = \frac{5}{2} \end{align*}
The solutions to the resulting system doesn’t require much work, sinc ethe first initial condition gives \(A\) directly and I used the second to calculate \(B\text{.}\) The specific solution is
\begin{equation*} y = 2 e^{-2t} \cos 4t + \frac{5}{2} e^{-2t} \sin 4t \end{equation*}
As you decrease the coefficient, \(\alpha\) remains unchanged since it only depends on the other two coefficients. Therefore, the rate of decay is unchanged. However, \(\beta = \frac{\sqrt{4ac - b^2}}{2a}\) does depend on the coefficient of \(y\text{,}\) which is \(c\) in the quadratic formula. As that coefficient becomes smaller and smaller, \(\beta\) also decreases. Since \(\beta\) measures frequency, decreasing \(beta\) means slower oscillations.
Critically damped oscillations happen when \(\beta = 0\text{,}\) or equivalently when \(b^2 - 4ac=0\text{.}\) In this system, since \(b=4\) and \(a = 1\text{,}\) the coefficient of \(y\) has to be \(c=4\text{.}\)
Exponential decay will happen below the \(c=4\) threshold, when \(b^2 -4ac\) is positive again and the solutions will be real exponentials.
In a harmonic system, you are changing the stiffness of the system. Decreasing the stiffness makes the system less responsive and less likely to have oscilations. For circuits, you are effect the. capacitance, which is (in some ways) analogous to stiffness in the circuit system.

Subsection 3.5.2 Undetermined Coefficients

Activity 3.5.3.

Solve this non-homogeneous SOCCLDE using undetermined coefficients.

\begin{align*} \amp y^{\prime \prime} + 3 y^\prime + 2y = e^{-7t} \amp \amp y(0) = 3 \amp \amp y^\prime(0) = 0 \end{align*}

Solution.

First I need the homogeneous solution. I start by solving the characteristic equation.

\begin{equation*} r^2 + 3r + 2 = 0 \implies r = -2, -1 \end{equation*}

The roots are real Therefore, the homogeneous solutions are exponential.

\begin{equation*} y_h = Ae^{-t} + Be^{-2t} \end{equation*}

My guess for the particular solution using undetermined coefficients some from the forcing term.

\begin{equation*} y_p = Ce^{-7t} \end{equation*}

I need the derivaitves of this particular solution.

\begin{gather*} y_p^{\prime} = -7Ce^{-7t} \\ y^{\prime \prime} = 49 Ce^{-7t} \end{gather*}

Then I put this into the differential equation and isolate like terms.

\begin{align*} 49Ce^{-7t} - 21Ce^{-7t} + 2Ce^{-7t} \amp = e^{-7t} \\ [49C - 21C + 2C] e^{-7t} = e^{-7t} \end{align*}

This produces a system of linear equations. (In this case, only one equation.)

\begin{equation*} 30C = 1 \implies C = \frac{1}{30} \end{equation*}

Now that I now the coefficients, I can write the particular solution and thus the general solutions by adding the homogeneous solution.

\begin{align*} y_p \amp = \frac{1}{30} e^{-7t} \\ y \amp = \frac{1}{30} e^{-7t} + Ae^{-t} + Be^{-2t} \end{align*}

I need the derivative of the general solution.

\begin{equation*} y^\prime = \frac{-7}{30} e^{-7t} - Ae^{-t} - 2B e^{-2t} \end{equation*}

Then I use the initial conditions. I get two equations, which I can turn into a linear system. I’ll solve that linear system by solving for \(A\) in the second equation and substiting that \(A\) in the first equation. That lets me solve for \(B\) and then use the second equation to finally find \(A\text{.}\)

\begin{align*} y(0) \amp = \frac{1}{30} + A + B = 3 \\ y^\prime (0) \amp = \frac{-7}{30} - A - 2B = 0 \implies A = \frac{-7}{30} - 2B\\ \frac{1}{30} - \frac{7}{30} - 2B + B \amp = 3 \implies B = \frac{-16}{5} \\ A \amp = \frac{-7}{30} - 2B = \frac{-7}{30} + \frac{32}{5} = \frac{37}{6} \end{align*}

Now I can write the specific solution to this IVP.

\begin{equation*} y_p = \frac{1}{30} e^{-7t} + \frac{37}{6} e^{-t} - \frac{16}{5} e^{-2t} \end{equation*}

Activity 3.5.4.

Solve this non-homogeneous SOCCLDE using undetermined coefficients.

\begin{align*} \amp y^{\prime \prime} + 2y^\prime + 10 t = \frac{t^2}{4} - t + 1 \amp \amp y(0) = 0 \amp \amp y^\prime(0) = 4 \end{align*}

Solution.

First I need the homogeneous solution. I start by solving the characteristic equation.

\begin{equation*} r^2 + 2r + 10 = 0 \implies r = -1 \pm \sqrt{-9} \end{equation*}

The roots are complex, with \(\alpha = -1\) and \(\beta = 3\text{.}\) Therefore, the homogeneous solutions are

\begin{equation*} y_h = Ae^{-t} \cos (3t) + Be^{-t} \sin (3t) \end{equation*}

My guess for the particular solution using undetermined coefficients some from the forcing term.

\begin{equation*} y_p = C_2 t^2 + C_1 t + C_0 \end{equation*}

I need the derivaitves of this particular solution.

\begin{align*} y_p^\prime \amp = 2C_2 t + C_1 \\ y_p^{\prime \prime} \amp = 2C_2 \end{align*}

Then I put this into the differential equation and isolate like terms.

\begin{equation*} [10C_2] t^2 + [4c_2 + 10C_1] t + [2C_2 + 2C_1 + 10C_0]t = \frac{1}{4} t^2 - t + 1 \end{equation*}

This produces a system of linear equations. I’ve simply asked a computer for the solutions to this system instead of doing the algebra myself.

\begin{align*} \amp 10C_2 = \frac{1}{4} \amp \amp 4C_2 + 10C_1 = -1 \amp \amp 2C_2 + 2C_1 + 10C_0 = 1 \\ \amp _2 = \frac{1}{40} \amp \amp C_1 = \frac{-11}{100} \amp \amp C_0 = \frac{117}{1000} \end{align*}

Now that I now the coefficients, I can write the particular solution and thus the general solutions by adding the homogeneous solution.

\begin{gather*} y_p = \frac{1}{40} t^2 - \frac{11}{100} t + \frac{117}{1000} \\ y = \frac{1}{40} t^2 - \frac{11}{100} t + \frac{117}{1000} + Ae^{-t} \cos(3t) + Be^{-t} \sin (3t) \end{gather*}

I need the derivative of the general solution. (The algebra is a bit lengthy, but I’ve done the derivatives and then groups like terms to make the derivative as easy to work with as I can)

\begin{equation*} y^\prime = \frac{1}{20}t - \frac{11}{100} + e^{-t} \left[ (-A + 3B) \cos (3t) + (-3A - B) \sin (3t) \right] \end{equation*}

Then I use the initial conditions. I get two equations, which I can turn into a linear system. The first equation, conveniently, only has \(A\text{,}\) so I can solve for \(A\) immediately. Then I can use that value to solve for \(B\) in the section equation.

\begin{align*} y(0) \amp = A + \frac{117}{1000} = 0 \implies A = \frac{-117}{1000}\\ y^\prime(0) \amp = (-A + 3B) - \frac{11}{100} = 4 \\ \frac{117}{1000} + 3B - \frac{11}{100} \amp = 4 \implies B = \frac{1331}{1000} \end{align*}

Now I can write the specific solution to this IVP.

\begin{equation*} y = \frac{1}{40} t^2 - \frac{11}{100}t + \frac{117}{1000} - \frac{117}{1000} e^{-t} \cos (3t) + \frac{1331}{1000} e^{-t} \sin(3t) \end{equation*}

Activity 3.5.5.

Solve this non-homogeneous SOCCLDE using undetermined coefficients.

\begin{align*} \amp y^{\prime \prime} + 4y^\prime + 8 y = 4 \sin (2t) \amp \amp y(0) = 0 \amp \amp y^\prime(0) = 0 \end{align*}

Solution.

First I need the homogeneous solution. I start by solving the characteristic equation.

\begin{equation*} r^2 + 4r + 8 = 0 \implies r = -2 \pm \sqrt{-4} \end{equation*}

The roots are complex with \(\alpha = -2\) and \(\beta = 2\text{.}\) Therefore, the homogeneous solutions are sinusoidal with exponential decay in amplitude.

\begin{equation*} y_h = Ae^{-2t} \cos (2t) + Be^{-2t} \sin (2t) \end{equation*}

My guess for the particular solution using undetermined coefficients some from the forcing term. Note that even those the forcing term and the homogeneous solution both involve the same sine funciton, I don’t have to multiply by \(t\) in my guess because the forcing term lacks the exponential decay of the homogeneous solution.

\begin{equation*} y_p = C \cos (2t) + D \sin (2t) \end{equation*}

I need the derivaitves of this particular solution.

\begin{align*} y_p^\prime \amp = -2C \sin (2t) + 2D cos (2t) \\ y_p^{\prime \prime} \amp = -4C \cos (2t) - 4D \sin (2t) \end{align*}

Then I put this into the differential equation and isolate like terms.

\begin{equation*} [4C + 8D] \cos (2t) + [4D - 8C] \sin (2t) = 4\sin (2t) \end{equation*}

This produces a system of linear equations. I haven’t shown the work for solving the system, though this is a pretty short isolate and replace system.

\begin{align*} \amp 4C + 8D = 0 \amp \amp 4D - 8C = 4 \\ \amp C = \frac{-2}{5} \amp \amp D = \frac{1}{5} \end{align*}

Now that I now the coefficients, I can write the particular solution and thus the general solutions by adding the homogeneous solution.

\begin{gather*} y_p = \frac{-=2}{5} \cos (2t) + \frac{1}{5} \sin (2t) \\ y_p = \frac{-=2}{5} \cos (2t) + \frac{1}{5} \sin (2t) + Ae^{-2t} \cos (2t) + Be^{-2t} \sin (2t) \end{gather*}

I need the derivative of the general solution. (I’ve skipped over some algebra steps and gathered together like to to make the derivative reasonable to work with.)

\begin{equation*} y^\prime = \frac{4}{5} \sin (2t) + \frac{2}{5} \cos (2t) + e^{-2t} \left[ (-2A + 2B) \cos (2t) + (-2A - 2B) \sin (2t) \right] \end{equation*}

Then I use the initial conditions. I get two equations, which I can turn into a linear system. The first equation of the linear system only involves \(A\text{,}\) so I can use it to solve for \(A\) directly. Then I can use that value of \(A\) to solve for \(B\) in the second system.

\begin{align*} y(0) \amp = A - \frac{2}{5} = 0 \implies A = \frac{2}{5} \\ y^\prime(0) \amp = (-2A + 2B) + \frac{2}{5} = 0 \implies B = \frac{-1}{5} \end{align*}

Now I can write the specific solution to this IVP.

\begin{equation*} y = \frac{-2}{5} \cos (2t) + \frac{1}{5} \sin (2t) + \frac{2}{5} e^{-2t} \cos (2t) - \frac{1}{5} e^{-2t} \sin (2t) \end{equation*}

Activity 3.5.6.

Solve this non-homogeneous SOCCLDE using undetermined coefficients.

\begin{align*} \amp y^{\prime \prime} + 6y^\prime + 18 y = 2e^{-3t} \cos (3t) \amp \amp y(0) = 1 \amp \amp y^\prime(0) = 0 \end{align*}

Solution.

First I need the homogeneous solution. I start by solving the characteristic equation.

\begin{equation*} r^2 + 6r + 18 = 0 \implies r = -3 \pm \sqrt{-9} \end{equation*}

The roots are complex with \(\alpha = -3\) and \(\beta = 3\text{.}\) Therefore, the homogeneous solutions are sinusoidal with exponentially decaying amplitude.

\begin{equation*} y_h = Ae^{-3t} \cos (3t) + Be^{-3t} \sin (3t) \end{equation*}

My guess for the particular solution using undetermined coefficients some from the forcing term. Here, the homogeneous solutions and the forcing term are the same, so I have to mulitply by \(t\) for the guess.

\begin{equation*} y_p = Cte^{-3t} \cos (3t) + Dte^{-3t} \sin (3t) \end{equation*}

I need the derivaitves of this particular solution. These derivatives are a little lengthy, but I did them by computer and wrote them as succinctly as I could.

\begin{align*} y_p^\prime \amp = [-3Ct + 3DT + C]e^{-3t} \cos (3t) + [-3Ct - 3Dt + D] e^{-3t} \sin (3t) \\ y^{\prime \prime}_p \amp = [-18Dt - 6C + 6D] e^{-3t} \cos (3t) + [18Ct - 6C + 6D] e^{-3t} \sin(3t) \end{align*}

Then I put this into the differential equation and isolate like terms. (Quite a bit of simplifying algebra is skipped over here.)

\begin{equation*} [6D]e^{-3t} \cos(3t) + [-6C + 12D] e^{-3t} \sin(3t) = 2e^{-3t} \cos (3t) \end{equation*}

This produces a system of linear equations. Even though the equations are quite complicated, the system is actually pretty reasonable: I solve for \(D\) in the first equation and use that to find \(C\) in the second.

\begin{align*} \amp 6D = 2 \amp \amp -6C + 12D = 0 \\ \amp C = \frac{2}{3} \amp \amp D = \frac{1}{3} \end{align*}

Now that I now the coefficients, I can write the particular solution and thus the general solutions by adding the homogeneous solution.

\begin{align*} y_p \amp = \frac{2}{3} te^{-3t} \cos (3t) + \frac{1}{3} te^{-3t} \sin (3t) \\ y \amp = \frac{2}{3} te^{-3t} \cos (3t) + \frac{1}{3} te^{-3t} \sin (3t) + Ae^{-3t} \cos (3t) + Be^{-3t} \sin (3t) \end{align*}

I need the derivative of the general solution. (Again, I’ve differentiated the complicated functions and grouped like terms to make this reasonable to work with.)

\begin{align*} y^\prime \amp = \left[ -3A + 3B + \frac{2}{3} \right] e^{-3t} \cos (3t) + \left[ -3A - 3B + \frac{1}{3} \right] e^{-3t} \sin (3t) \\ \amp - e^{-3t} t \cos (3t) - 3e^{-3t} t \sin (3t) \end{align*}

Then I use the initial conditions. I get two equations, which I can turn into a linear system. I’ll solve that linear system by solving for \(A\) in the first equation and using that to solve for \(B\) in the second.

\begin{align*} y(0) \amp = A = 1 \\ y^\prime(0) \amp = -3A + 3B + \frac{2}{3} = 0 \implies B = \frac{1}{9} \end{align*}

Now I can write the specific solution to this IVP.

\begin{equation*} y = \frac{2}{3} te^{-3t} \cos (3t) + \frac{1}{3} te^{-3t} \sin (3t) + e^{-3t} \cos (3t) + \frac{1}{9} e^{-3t} \sin (3t) \end{equation*}

Subsection 3.5.3 Variation of Parameters

Activity 3.5.7.

Solve this SOCCLDE using variation of parameters. (You can leave the resulting integrals if they are unsolvable.)

\begin{equation*} y^{\prime \prime} + 9y^\prime + 12y = \ln (t+3) \end{equation*}

Solution.

First I need to solve the homogeneous equation. I write and solve the characteristic equation.

\begin{equation*} r^2 + 9r + 12 = 0 \implies r = \frac{-9 \pm \sqrt{33}}{2} \end{equation*}

The roots are real. Therefore, the homogeneous solutions are exponential functions.

\begin{equation*} y_h = A e^{ \left( \frac{-9 + \sqrt{33}}{2} \right) t} + B e^{ \left( \frac{-9 - \sqrt{33}}{2} \right) t} \end{equation*}

Then I need to calculate the Wronskian. For this I need the derivatives of the two linearly independent homogeneous solutions.

\begin{align*} y_1 \amp = e^{ \left( \frac{-9 + \sqrt{33}}{2} \right) t} \\ y_2 \amp = e^{ \left( \frac{-9 - \sqrt{33}}{2} \right) t} \\ y_1^\prime \amp = \left( \frac{-9 + \sqrt{33}}{2} \right) e^{ \left( \frac{-9 + \sqrt{33}}{2} \right) t} \\ y_2^\prime \amp = \left( \frac{-9 - \sqrt{33}}{2} \right) e^{ \left( \frac{-9 - \sqrt{33}}{2} \right) t} \\ W \amp = e^{ \left( \frac{-9 + \sqrt{33}}{2} \right) t} \left( \frac{-9 - \sqrt{33}}{2} \right) e^{ \left( \frac{-9 - \sqrt{33}}{2} \right) t} \\ \amp - e^{ \left( \frac{-9 - \sqrt{33}}{2} \right) t} \left( \frac{-9 + \sqrt{33}}{2} \right) e^{ \left( \frac{-9 + \sqrt{33}}{2} \right) t} = -\sqrt{33} e^{ 24t} \end{align*}

Then I can calculate the two ‘varied parameter’ function using the homogeneous solutions and the Wronskian.

\begin{align*} u_1 \amp = -\int \frac{y_2 f}{W} dt = - \int \frac{ e^{ \left( \frac{-9 - \sqrt{33}}{2} \right) t} \ln (t+3}{ -\sqrt{33} e^{24t}} dt \\ \amp = \frac{1}{\sqrt{33}} \int e^{ \left( \frac{-57 - \sqrt{33}}{2} \right) t} \ln (t+3) dt\\ u_2 \amp = \int \frac{y_1 f}{W} dt = \int \frac{ e^{ \left( \frac{-9 + \sqrt{33}}{2} \right) t} \ln (t+3)}{ -\sqrt{33} e^{24t}} dt = \frac{-1}{\sqrt{33}} \int e^{ \left( \frac{-57 + \sqrt{33}}{2} \right) t} \ln (t+3) dt \end{align*}

These integrals are not solvable, so I just write the general solutoin in terms of these integrals.

\begin{align*} y \amp = A e^{ \left( \frac{-9 + \sqrt{33}}{2} \right) t} + B e^{ \left( \frac{-9 - \sqrt{33}}{2} \right) t} + \frac{1}{\sqrt{33}} e^{ \left( \frac{-9 + \sqrt{33}}{2} \right) t} \int e^{ \left( \frac{-57 - \sqrt{33}}{2} \right) t} \ln (t+3) dt \\ \amp - \frac{1}{\sqrt{33}} e^{ \left( \frac{-9 - \sqrt{33}}{2} \right) t} \int e^{ \left( \frac{-57 + \sqrt{33}}{2} \right) t} \ln (t+3) dt \end{align*}

Activity 3.5.8.

Solve this SOCCLDE using variation of parameters. (You can leave the resulting integrals if they are unsolvable.)

\begin{equation*} y^{\prime \prime} + 10y^\prime + 5y = \arctan t \end{equation*}

Solution.

First I need to solve the homogeneous equation. I write and solve the characteristic equation.

\begin{equation*} r^2 + 10r + 5 = 0 \implies r = -5 \pm \sqrt{20} \end{equation*}

The roots are complex with \(\alpha = -5\) and \(\beta = \sqrt{20}\) Therefore, the homogeneous solutions are sinusoidal with exponential decay.

\begin{equation*} y_h = A e^{-5t} \cos (\sqrt{20} t) + B e^{-5t} \sin (\sqrt{20} t ) \end{equation*}

Then I need to calculate the Wronskian. For this I need the derivatives of the two linearly independent homogeneous solutions.

\begin{align*} y_1 \amp = e^{-5t} \cos ( \sqrt{20} t) \\ y_2 \amp = e^{-5t} \sin ( \sqrt{20} t) \\ y_1^\prime \amp = -5e^{-5t} \cos (\sqrt{20} t) - e^{-5t} \sin (\sqrt{20} t) \\ y_2^\prime \amp = -5e^{-5t} \sin (\sqrt{20} t) + e^{-5t} \cos (\sqrt{20} t) \\ W \amp = e^{-10 t} \left[ \cos^2 (\sqrt{20}t) -5 \cos (\sqrt{20} t) \sin ( \sqrt{20} t) \right] \\ \amp - e^{-10t} \left[ -5 \sin (\sqrt{20} t) \cos (\sqrt{20} t) - \sin^2 ( \sqrt{20} t) \right] = e^{-10t} \end{align*}

Then I can calculate the two ‘varied parameter’ function using the homogeneous solutions and the Wronskian.

\begin{gather*} u_1 = -\int \frac{y_2 f}{W} dt = -\int \frac{ e^{-5t} \sin (\sqrt{20} t) \arctan t}{e^{-10t}} dt \\ = - \int e^{5t} \sin (\sqrt{20}t) \arctan t dt \\ u_2 = \int \frac{y_1 f}{W} dt = \int \frac{ e^{-5t} \cos (\sqrt{20} t) \arctan t}{e^{-10 t}} dt \\ = \int e^{5t} \cos (\sqrt{20} t) \arctan t dt \end{gather*}

These integrals are not solvable, so I just write the general solutoin in terms of these integrals.

\begin{align*} y \amp = A e^{-5t} \cos (\sqrt{20} t) + B e^{-5t} \sin (\sqrt{20} t )\\ \amp - e^{-5t} \cos ( \sqrt{20} t) \int e^{5t} \sin (\sqrt{20}t) \arctan t dt \\ \amp + e^{-5t} \sin ( \sqrt{20} t) \int e^{5t} \cos (\sqrt{20} t) \arctan t dt \end{align*}

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