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

(8)