Method of Variation of Parameters: Higher Order Nonhom. Diff. Eqs.

# The Method of Variation of Parameters for Higher Order Nonhomogeneous Differential Equations

Recall from The Method of Variation of Parameters page, we were able to solve many different types of second order linear nonhomogeneous differential equations with constant coefficients by first solving for the solution to the corresponding linear homogeneous differential equation for $y_h(t) = Cy_1(t) + Dy_2(t)$ (where $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions), and then substituting the constants $C$ and $D$. We then assumed that a particular solution for the nonhomogeneous differential equation was of the form $Y(t) = u_1(t)y_1(t) + u_2(t)y(t)$, and we solved the following system of linear equations for $u_1'(t)$ and $u_2'(t)$:

(1)
\begin{align} \quad \left\{\begin{matrix} u_1'(t)y_1(t) + u_2'(t)y_2(t) = 0 \\ u_1'(t)y_1'(t) + u_2'(t)y_2'(t) = g(t) \end{matrix}\right. \end{align}

After we obtained the functions $u_1'(t)$ and $u_2'(t)$, we integrated the results to obtain $u_1(t)$ and $u_2(t)$, and then plugged them into the formula $Y(t) = u_1(t)y_1(t) + u_2(t)y_2(t)$ for a particular solution (the existence and uniqueness of a solution is guaranteed from $y_1(t)$ and $y_2(t)$ forming a fundamental set of solutions). The general solution to our second order nonhomogeneous differential equation was then:

(2)
\begin{align} \quad y(t) = y_h(t) + Y(t) \quad y(t) = Cy_1(t) + Dy_2(t) + u_1(t)y_1(t) + u_2(t)y_2(t) \end{align}

As well will now see - the method of variation of parameters can also be applied to higher order differential equations. The process can be derived similarly. Suppose that we have a higher order differential equation of the following form:

(3)
\begin{align} \quad \frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}}{dt^{n-1}} + ... + p_n(t) y = g(t) \end{align}

We first solve the corresponding homogeneous differential equation to get $y_h(t) = C_1 y_1(t) + C_2y_2(t) + ... + C_ny_n(t)$. The functions $y_1(t)$, $y_2(t)$, …, $y_n(t)$ form a fundamental set. Assume a particular solution to the nonhomogeneous differential equation is of the form:

(4)
\begin{align} \quad Y(t) = u_1(t)y_1(t) + u_2(t)y_2(t) + ... + u_n(t)y_n(t) \end{align}

We then solve the following system of equations for the functions $u_1'(t)$, $u_2'(t)$, …, $u_n'(t)$.

(5)
\begin{align} \quad \left\{\begin{matrix} u_1'(t)y_1(t) + u_2'(t)y_2(t) + ... + u_n'(t)y_n(t) = 0 \\ u_1'(t)y_1'(t) + u_2'(t)y_2'(t) + ... + u_n'(t)y_n'(t) = 0 \\ \vdots \\ u_1'(t)y_1^{(n-1)}(t) + u_2'(t)y_2^{(n-1)}(t) + ... + u_n'(t)y_n^{(n-1)}(t) = g(t) \end{matrix}\right. \end{align}

Once again, a unique solution is guaranteed since $y_1(t)$, $y_2(t)$, …, $y_n(t)$ form a fundamental set of solutions implies the Wronskian $W(y_1, y_2, ..., y_n)$ is nonzero. Furthermore, it should be noted that the system above can be solved for with row reduction (if the process is simple) or more commonly by applying Cramer's rule once again.

We then integrate each of the functions $u_1'(t)$, $u_2'(t)$, …, $u_n'(t)$ to obtain $u_1(t)$, $u_2(t)$, …, $u_n(t)$. Lastly, we obtain that $Y(t) = u_1(t)y_1(t) + u_2(t)y_2(t) + ... + u_n(t)y_n(t)$ as our particular solution.

We will now look at an example of using the method of variation of parameters for higher order nonhomogeneous differential equations.

## Example 1

Solve the third order linear nonhomogeneous differential equation $y''' + y' = \tan t$ using the method of variation of parameters.

For the corresponding homogeneous differential equation $y''' + y' = 0$ we have that the characteristic equation is $r^3 + r = 0$ which can be factored as $r(r^2 + 1) = 0$ and so $r = 0$ or $r = \pm i$. The solution to the corresponding homogeneous differential equation is therefore:

(6)
\begin{align} \quad y(t) = P + Q\cos t + R \sin t \end{align}

Furthermore, $y_1 = e^{0t} = 1$, $y_2 = e^{0t} \cos t= \cos t$, and $y_3 = e^{0t} \sin t = \sin t$.

Now let's try to find a particular solution to this differential equation. We have that:

(7)
\begin{align} \quad y_p = u_1y_1 + u_2y_2 + u_3y_3 \\ \quad y_p = u_1 + u_2 \cos t + u_3 \sin t \end{align}

Thus we want to solve the following system of equations:

(8)
\begin{align} \quad u_1' (1) + u_2' \cos t + u_3' \sin t = 0 \\ \quad u_1' (0) + u_2' (-\sin t) + u_3' (\cos t) = 0 \\ \quad u_1' (0) + u_2' (\cos t) + u_3' (\sin t) = \tan t \end{align}

We will use Cramer's rule in order to solve for $u_1'$, $u_2'$, and $u_3'$. We first find the corresponding Wronskian:

(9)
\begin{align} \quad W = \begin{bmatrix} 1 & \cos t & \sin t \\ 0 & -\sin t & \cos t \\ 0 & -\cos t & -\sin t \end{bmatrix} = \begin{bmatrix} -\sin t & \cos t \\ -\cos t & -\sin t \end{bmatrix} = 1 \end{align}

Now we have that:

(10)
\begin{align} \quad u_1' = \frac{\begin{bmatrix} 0 & \cos t & \sin t \\ 0 & -\sin t & \cos t \\ \tan t & -\cos t & -\sin t \end{bmatrix}}{1} = \tan t \begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix} = \tan t \end{align}
(11)
\begin{align} \quad u_2' = \frac{\begin{bmatrix} 1 & 0 & \sin t \\ 0 & 0 & \cos t \\ 0 & \tan t & -\sin t \end{bmatrix}}{1} = - \begin{bmatrix} 1 & 0 & \sin t \\ 0 & \tan t & -\sin t \\ 0 & 0 & \cos t \end{bmatrix} = -\tan t \cos t = -\frac{\sin t}{\cos t} \cos t = -\sin t \end{align}
(12)
\begin{align} \quad u_3' = \frac{\begin{bmatrix} 1 & \cos t & 0 \\ 0 & -\sin t & 0 \\ 0 & -\cos t & \tan t \end{bmatrix}}{1} = \begin{bmatrix} -\sin t & 0 \\ -\cos t & \tan t \end{bmatrix} = -\sin t \tan t = -\frac{\sin^2 t}{\cos t} \end{align}

We will now integrate $u_1'$, $u_2'$, and $u_3'$ to get:

(13)
\begin{align} \quad \int u_1'(t) \: dt = \int \tan t \: dt = \ln \mid \sec t \mid + C \end{align}
(14)
\begin{align} \quad \int u_2'(t) \: dt = \int - \sin t \: dt = \cos t + D \end{align}
(15)
\begin{align} \quad \int u_3'(t) \: dt = \int -\frac{\sin^2 t}{\cos t} \: dt = \int \frac{\cos^2 t - 1}{\cos t} \: dt = \int \left ( \cos t - \sec t \right ) \: dt = \sin t + \ln \mid \sec t + \tan t \mid + E \end{align}

Thus we have that:

(16)
\begin{align} \quad y(t) = \left ( \ln \mid \sec t \mid + C \right ) (1) + \left ( \cos t + D \right ) \cos t + \left ( \sin t + \ln \mid \sec t + \tan t \mid + E \right ) \sin t \end{align}