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:

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

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