Problems from Homogeneous linear equations of order 2 with non constant coefficients

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:

$x_2=x_1\cdot u$

$x_2^{\prime }=x_1^{\prime }\cdot u+x_1\cdot u^{\prime }$

$x_2^{″}=x_1^{″}\cdot u+x_1^{\prime }\cdot u^{\prime }+x_1^{\prime }\cdot u^{\prime }+x_1\cdot u^{″}=x_1^{″}\cdot u+2x_1^{\prime }\cdot u^{\prime }+x_1{u}^{″}$

Let's designate this as a solution:

$$4t^2{x}_2^{″}+x_2=4t^2(x_1^{″}\cdot u+2x_1^{\prime }\cdot u^{\prime }+x_1{u}^{″})+x_1\cdot u=$$ $$=u^{″}(4t^2\cdot x_1)+u^{\prime }(8t^2\cdot x_1^{\prime })+u(4t^2\cdot x_1^{″}+x_1)=0$$ Notice that the last term equals zero, since $x_1(t)$ is a solution. Introducing the change $w(t)=u^{\prime }(t)$, we obtain a linear ODE in $w(t)$: $$w^{\prime }(4t^2\cdot x_1)+w(8t^2\cdot x_1^{\prime })=0\Rightarrow {\displaystyle \frac{dw}{dt}}={\displaystyle \frac{-w(8t^2\cdot x_1^{\prime })}{4t^2\cdot x_1}}\Rightarrow$$ $$\Rightarrow {\displaystyle \frac{dw}{w}}={\displaystyle \frac{-2x_1^{\prime }}{x_1}}dt\Rightarrow \int {\displaystyle \frac{dw}{w}}=\int {\displaystyle \frac{-2x_1^{\prime }}{x_1}}dt\Rightarrow$$ $$\Rightarrow \ln |w(t)|=-2\cdot \ln |x_1(t)|=\ln (|x_1(t){|}^{-2})\Rightarrow$$ $$\Rightarrow w(t)={\displaystyle \frac{1}{x_1(t{)}^2}}={\displaystyle \frac{1}{t\cdot (\ln (t){)}^2}}$$ We now compute the primitive function of $w(t)$ to find $u(t)$: $$u(t)=\int {\displaystyle \frac{1}{t\cdot (\ln (t){)}^2}}dt={\displaystyle \frac{1}{\ln (t)}}$$ Finally: $$x_2(t)=x_1(t)\cdot u(t)=\sqrt{t}\cdot \ln (t)\cdot {\displaystyle \frac{1}{\ln (t)}}=\sqrt{t}$$ 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:

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

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