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:
Determine a second solution to the DE \( 2t^2y^{''} +ty^{'} - 3y = 0\) given \(y_1 = t^{3/2}\) is a solution.
This is a second-order, homogenous, linear equation so we can use the reduction of order techqiue. First, put the equation in standard form:
\(y^{''} + \cfrac{1}{2t}y^{'} - \cfrac{3}{2t^2}y = 0\)
Now, identify \(P(t)\) and calculate factor \(e^ {-\int P(t)dt}\):
\(P(t) = \cfrac{1}{2t}\)
\(e^{-\int P(t) dt} = e^{-\int \cfrac{1}{2t} dt} = e^{-1/2 \ln t} = t^{-1/2}\)
Next, divide by the square of the first solution:
\( \cfrac{e^ {-\int P(t) dt}} {y_1(t)^2} = \cfrac{t^{-1/2}} {(t^{3/2})^2} = \cfrac{t^{-1/2}}{t^3} = t^{-7/2}\)
Next, take the integral of this term:
\(v(t) = \int {\cfrac{e^ {-\int P(t)dt}} {y_1(t)^2}}dt = \int { t^{-7/2} dt} = \frac{-2}{5} t^{-5/2} \)
Finally, the second solution is:
\( y_2 = v(t)y_1 = \frac{-2}{5} t^{-5/2} \cdot t^{3/2} = \frac{-2}{5} t^{-1} \)
The general solution would be:
\(y = c_1t^{3/2} + c_2 t^{-1}\)