Solve the equation $$4t^2x''+x=0$$ knowing that one solution is $$x_1(t)=\sqrt{t} \cdot \ln t$$
Development:
This is an homogeneous 2nd order ODE with non constant coefficients. As we already know one solution, the method of reduction of order will give us another linearly independent solution $$x_2(t)$$, so that we will obtain two linearly independent solutions that will solve the ODE, since any solution can be written: $$$x(t)=C_1\cdot x_1(t)+C_2\cdot x_2(t)$$$
We look for the second solution of the form: $$x_2(t)=x_1(t)\cdot u(t)$$.
We compute:
$$x_2=x_1\cdot u$$
$$x_2'=x_1'\cdot u+x_1\cdot u'$$
$$x_2''=x_1''\cdot u+x_1'\cdot u'+x_1'\cdot u'+x_1\cdot u''=x_1''\cdot u+2x_1'\cdot u' + x_1 u''$$
Let's designate this as a solution:
$$$4t^2x_2''+x_2=4t^2(x_1''\cdot u +2x_1'\cdot u'+x_1 u'')+x_1\cdot u=$$$ $$$=u''(4t^2\cdot x_1)+u'(8t^2\cdot x_1')+u(4t^2\cdot x_1''+x_1)=0$$$ Notice that the last term equals zero, since $$x_1(t)$$ is a solution. Introducing the change $$w(t)=u'(t)$$, we obtain a linear ODE in $$w(t)$$: $$$w'(4t^2\cdot x_1)+w(8t^2\cdot x_1')=0 \Rightarrow \dfrac{dw}{dt}=\dfrac{-w(8t^2\cdot x_1')}{4t^2\cdot x_1} \Rightarrow$$$ $$$ \Rightarrow \dfrac{dw}{w}=\dfrac{-2x_1'}{x_1}dt \Rightarrow \int\dfrac{dw}{w}=\int\dfrac{-2x_1'}{x_1}dt \Rightarrow$$$ $$$\Rightarrow \ln|w(t)|=-2\cdot\ln|x_1(t)|=\ln(|x_1(t)|^{-2}) \Rightarrow$$$ $$$\Rightarrow w(t)=\dfrac{1}{x_1(t)^2}=\dfrac{1}{t\cdot(\ln(t))^2}$$$ We now compute the primitive function of $$w (t)$$ to find $$u(t)$$: $$$u(t)=\int\dfrac{1}{t\cdot(\ln(t))^2}dt=\dfrac{1}{\ln(t)}$$$ Finally: $$$x_2(t)=x_1(t)\cdot u(t)=\sqrt{t}\cdot\ln(t)\cdot\dfrac{1}{\ln(t)}=\sqrt{t}$$$ Therefore any solution will be written as: $$$x(t)=C_1\cdot x_1(t)+C_2\cdot x_2(t)=C_1(t)\cdot\sqrt{t}\cdot\ln(t)+C_2(t)\cdot\sqrt{t}$$$
Solution:
$$x(t)=C_1(t)\cdot\sqrt{t}\cdot\ln(t)+C_2(t)\cdot\sqrt{t}$$