Reduction of Order - Second Order Lin. Hom. Diff. Eqs. Examples 1

Reduction of Order on Second Order Linear Homogeneous Differential Equations Examples 1

Recall from the Reduction of Order on Second Order Linear Homogenous Differential Equations page that if we have a second order linear homogeneous differential equation $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ and if $y = y_1(t)$ is a nonzero solution to this differential equation then suppose that $y = v(t) y_1(t)$ is also a solution to this differential equation. We then saw that by taking the first and second derivatives, $y'$ and $y''$, then we could reduce our differential equation to:

(1)

\begin{align} \quad (2y_1'(t) + p(t)y_1(t))v'(t) + y_1(t)v''(t) = 0 \end{align}

But this is merely a first order differential equation of the function $v'(t)$ and so we can use techniques of first order differential equations to solve for $v(t)$ and then for $y_2 = v(t) y_1(t)$ and possibly form a fundamental set of solutions to the original differential equations.

Let's look at some examples of reduction of order on second order linear homogeneous differential equations.

Example 1

Find a second solution to the differential equation $t^2 \frac{d^2y}{dt^2} + 2t \frac{dy}{dt} - 2y = 0$ for $t > 0$ given that $y_1(t) = t$ is a solution.

Let $y = v(t) y_1(t) = tv(t)$. If we differentiate this function, we get that $y' = tv'(t) + v(t)$, and if we differentiate again, we get that $y''(t) = tv''(t) + v'(t) + v'(t) = tv''(t) + 2v'(t)$. If we plug this into our differential equation then we have that:

(2)

\begin{align} \quad t^2 \frac{d^2 y}{dt^2} + 2t \frac{dy}{dt} - 2y = 0 \\ \quad t^2 \left [ tv''(t) + 2v'(t) \right ] + 2t \left [ tv'(t) + v(t) \right ] - 2 \left [ tv(t) \right ] = 0 \\ \quad t^3 v''(t) + 2t^2 v'(t) + 2t^2 v'(t) + 2t v(t) - 2t v(t) = 0 \\ \quad t^3 v''(t) + 4t^2 v'(t) = 0 \end{align}

We can now solve this differential equation using techniques for solving first order differential equations. Let $u(t) = v'(t)$. Then we obtain the following first order differential equation:

(3)

\begin{align} \quad t^3 u'(t) + 4t^2 u(t) = 0 \\ \quad u'(t) + \frac{4}{t} u(t) = 0 \end{align}

Let's solve this differential equation by using integrating factors. Let $\mu (t) = e^{ \int \frac{4}{t} \: dt} = e^{4\ln t} = e^{\ln t^4} = t^4$. Now let's multiply both sides of the differential equation above by $\mu (t)$ to get that:

(4)

\begin{align} \quad \mu(t) u'(t) + \mu(t) \frac{4}{t} u(t) = 0 \\ \quad t^4 u'(t) + 4t^3 u(t) = 0 \\ \quad \frac{d}{dt} (t^4 u(t)) = 0 \\ \quad \int \frac{d}{dt} (t^4 u(t)) \: dt = \int 0 \: dt \\ \quad t^4 u(t) = C \\ \quad u(t) = Ct^{-4} \end{align}

Now note that $u(t) = v'(t)$ so $v'(t) = Ct^{-4}$. Integrating both sides of this equation and we get that $v(t) = -C\frac{t^{-3}}{3} = -\frac{C}{3t^3}$. Therefore:

(5)

\begin{align} \quad y = v(t) y_1(t) \\ \quad y = -\frac{C}{3t^3} \cdot t \\ \quad y = -\frac{C}{3t^2} \end{align}

Note that $-\frac{C}{3}$ is just a constant, and more simply, $y_2 = \frac{1}{t^2}$ is another solution to this differential equation.

Example 2

Find a second solution to the differential equation $t \frac{d^2y}{dt^2} - \frac{dy}{dt} + 4t^3 y = 0$ for $t > 0$ given that $y_1(t) = \sin t^2$

First let's rewrite our differential equation by diving by $t$ to get:

(6)

\begin{align} \quad \frac{d^2 y}{dt} - \frac{1}{t} \frac{dy}{dt} + 4t^2 y = 0 \end{align}

Let $y = v(t) y_1(t) = v(t) \sin t^2$. Instead of computing the first and second derivatives of $y$, let's instead use the formula:

(7)

\begin{align} \quad (2y_1'(t) + p(t)y_1(t))v'(t) + y_1(t) v''(t) = 0 \\ \quad \left (4t \cos t^2 -\frac{1}{t}\sin t^2 \right ) v'(t) + \sin t^2 v''(t) = 0 \end{align}

Now let $u(t) = v'(t)$ and so we obtain the following first order differential equation:

(8)

\begin{align} \quad \left (4t \cos t^2 -\frac{1}{t}\sin t^2 \right ) u(t) + \sin t^2 u'(t) = 0 \end{align}