# Reduction of Order

Given one solution to a DE, we can find another solution that is linearly independent using a technique refered to as Reduction of Order. The idea is to find a function $$v(t)$$ such that the second solution is $$y_2 = v(t)y_1$$.

Consider a second-order homogenous differential equation in standard form:

$$y^{''} + P(t)y^{'} + Q(t)y = 0$$

We can substitute $$y_2$$ along with its first and second derivatives (using chain rule) to obtain:

$$y_1 v^{''} + (2y_1^{'} + P(t) y_1)v^{'} + (y_1^{''} + P(t)y_1^{'} + Q(t)y_1)v = 0$$

Notice that the coefficient on $$v$$ is just the original DE which is equal to 0. The equation becomes:

$$y_1 v^{''} + (2y_1^{'} + P(t) y_1)v^{'} = 0$$

If we make the substitution $$u = v^{'}$$ we obtain a first order DE that can be solved using an integration factor or simply by integrating if it is separable.

$$y_1 u^{'} + (2y_1^{'} + P(t) y_1) u = 0$$

Once we solve for $$u(t)$$, we can integrate to solve for $$v(t)$$. The second, linearly independent solution to the DE is $$y_2 = u(t)y_1$$.

The steps to obtain the function $$v(t)$$ are:

1. Write the equation in standard form $$\cfrac{d^2y}{dt^2} + P(t) \cfrac{dy}{dt} + Q(t)y = 0$$.
2. Find the factor $$e^ {-\int P(t) dt}$$.
3. Find the factor $$\cfrac{e^ {-\int P(t) dt}} {y_1(t)^2}$$.
4. Solve $$v(t) = \int {\cfrac{e^ {-\int P(t)dt}} {y_1(t)^2}}dt$$.
5. The second solution is $$y_2 = v(t)y_1$$
6. The general solution is $$y = c_1y_1 + c_2y_2$$

Determine a second solution to the DE $$2t^2y^{''} +ty^{'} - 3y = 0$$ given $$y_1 = t^{3/2}$$ is a solution.