Homogeneous linear equations of order 2 with non constant coefficients

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: $$x(t)=C_1\cdot x_1(t)+C_2\cdot x_2(t)=C_1(t)\cdot\sqrt{t}\cdot\ln(t)+C_2(t)\cdot\sqrt{t}$$