Here is a general formula when you are given two solutions \(\displaystyle y_{1}(t), \;\ y_{2}(t)\)
If you have a second order equation, \(\displaystyle y''+a_{1}(t)y'+a_{2}y=b(t)\)
We have \(\displaystyle W_{1}(t)=-y_{2}(t), \;\ W_{2}(t)=y_{1}(t)\) for any pair of fundamental pair of solutions \(\displaystyle y_{1}(t), \;\ y_{2}(t)\) of a
homogeneous equation. If W(t) is their Wronskian, then the solution y(t) satisfying \(\displaystyle t(t_{0})=y_{0}, \;\ y'(t_{0})=y'_{0}\) is given
by, brace yourself, \(\displaystyle y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)-y_{1}(t)\int_{t_{0}}^{t}\frac{b(s)y_{2}(s)}{W(s)}ds+y_{2}(t)\int_{t}^{t_{0}}\frac{b(s)y_{1}(s)}{W(s)}ds\). s is a dummy variable, so to speak.
Where \(\displaystyle c_{1}, \;\ c_{2}\) are constants chosen so that \(\displaystyle {\phi}(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)\) satisfies the initial conditions.
I know....i know...it looks monstrous. But it is not as bad as it looks. You are given the info to solve. It is a general form to find the other solution.
For instance, suppose we have \(\displaystyle y_{1}(t)=\sqrt{t}, \;\ y_{2}(t)=\sqrt{t}ln(t)\) are a solution of the fundamental system of solutions of the equation
\(\displaystyle y''+\frac{1}{4t^{2}}y=0\) and W(t)=1.
Using the expression and letting \(\displaystyle t_{0}=1\), we can find out that the solution y(t) satisfying y(1)=1 and y'(1)=3/2 of
\(\displaystyle y''+\frac{1}{4t^{2}}y=t^{\frac{3}{2}}\) is given by \(\displaystyle y(t)=\frac{8}{9}\sqrt{t}+\frac{2}{3}\sqrt{t}ln(t)+\frac{1}{9}t^{\frac{7}{2}}\)
This is what I use to solve many reduction of order problems. It can make them easier.