Find general solutions to the nonhomogeneous Cauchy-Euler equations using a variety of parameters.

${\mathbf{t}}^{\mathbf{2}}\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{tz}\mathbf{\text{'}}\mathbf{+}\mathbf{z}\mathbf{=}\mathbf{t}\left(\mathbf{1}\mathbf{+}\frac{\mathbf{3}}{\mathbf{lnt}}\right)$

First, we will find a homogeneous solution to the given equation. To do so, we will take substituting$\mathrm{t}={\mathrm{e}}^{\mathrm{x}}$ which will transform the given equation into a constant coefficient equation. In Problem 23 it is shown that $\mathrm{tz}\text{'}=\frac{\mathrm{dZ}}{\mathrm{dx}},{\mathrm{t}}^2\mathrm{z}\text{'}\text{'}=\frac{{\mathrm{d}}^2\mathrm{Z}}{{\mathrm{dx}}^2}-\frac{\mathrm{dZ}}{\mathrm{dx}}$, where$\mathrm{Z}\left(\mathrm{x}\right)=\mathrm{z}\left({\mathrm{e}}^{\mathrm{x}}\right)$

Substituting this we transform the equation ${\mathrm{at}}^2\mathrm{z}\text{'}\text{'}+\mathrm{btz}\text{'}+\mathrm{cz}=\mathrm{f}\left(\mathrm{t}\right)$to

$\mathrm{a}\frac{{\mathrm{d}}^2\mathrm{Z}}{{\mathrm{dx}}^2}+(\mathrm{b}-\mathrm{a})\frac{\mathrm{dZ}}{\mathrm{dx}}+\mathrm{cZ}=\mathrm{f}\left({\mathrm{e}}^{\mathrm{x}}\right)$

The coefficient multiplying ${\mathrm{t}}^2\mathrm{z}\text{'}\text{'}\left(\mathrm{t}\right)$is $\mathrm{a}=1$, the one multiplying $\mathrm{tz}\text{'}\left(\mathrm{t}\right)$is $\mathrm{b}=-1$, the one multiplying $\mathrm{y}\left(\mathrm{t}\right)$is $\mathrm{c}=1$and the non-homogeneous part of the given equation is $\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}\left(1+\frac{3}{\mathrm{lnt})}\right)$. Therefore, the given equation will transform by this substitution into

$\begin{array}{rcl}1\times \frac{{\mathrm{d}}^2\mathrm{Z}}{\mathrm{dx}}+(-1-1)\frac{\mathrm{dZ}}{\mathrm{dx}}+1\times \mathrm{Z}& =& {\mathrm{e}}^{\mathrm{x}}\left(1+\frac{3}{{\mathrm{lne}}^{\mathrm{x}}}\right)\\ \frac{{\mathrm{d}}^2\mathrm{Z}}{\mathrm{dx}}-2\frac{\mathrm{dZ}}{\mathrm{dx}}+\mathrm{Z}& =& {\mathrm{e}}^{\mathrm{x}}\left(1+\frac{3}{\mathrm{x}}\right)\end{array}$

One will solve the corresponding homogeneous equation $\frac{{\mathrm{d}}^2\mathrm{Z}}{\mathrm{dx}}-2\frac{\mathrm{dZ}}{\mathrm{dx}}+\mathrm{Z}=0$. The auxiliary equation is ${\mathrm{r}}^2-2\mathrm{r}+1=(\mathrm{r}-1{)}^2$and its roots are ${\mathrm{r}}_{1,2}=1$. Since we have a double root to the auxiliary equation, the homogeneous solution is ${\mathrm{Z}}_{\mathrm{h}}\left(\mathrm{x}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2\mathrm{x}{3}^{\mathrm{x}}$

Expressing this in terms of t, the homogeneous solution to the given differential equation is:

$$\begin{gathered} {\mathrm{z}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{lnt}}+{\mathrm{c}}_2{\mathrm{lnte}}^{\mathrm{lnt}} \\ {\mathrm{z}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{t}+{\mathrm{c}}_2\mathrm{tlnt} \end{gathered}$$

To find a particular solution, we will apply the method of variation of parameters. For two linearly solutions of the homogeneous equation we will take${\mathrm{z}}_1\left(\mathrm{t}\right)=\mathrm{t}$ and${\mathrm{z}}_2\left(\mathrm{t}\right)=\mathrm{tlnt}$

But before we apply this method, we need to transform the given equation so that the coefficient multiplying$\mathrm{z}\text{'}\text{'}$ is 1 and in the given equation this coefficient is a function ${\mathrm{t}}^2$.

To transform it to the required form we will divide the given equation by ${\mathrm{t}}^2$.

$\mathrm{z}\text{'}\text{'}-\frac{1}{\mathrm{t}}\mathrm{z}\text{'}+\frac{1}{{\mathrm{t}}^2}\mathrm{z}=\frac{\mathrm{lnt}+3}{\mathrm{tlnt}}$

One can assume that a particular solution has a form of ${\mathrm{z}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{z}}_1\left(\mathrm{t}\right)+{\mathrm{v}}_2\left(\mathrm{t}\right){\mathrm{z}}_2\left(\mathrm{t}\right)$, where ${\mathrm{v}}_1\left(\mathrm{t}\right)$and ${\mathrm{v}}_2\left(\mathrm{t}\right)$are functions that could be found from ${\mathrm{v}}_1\left(\mathrm{t}\right)=\int \frac{-\mathrm{g}\left(\mathrm{t}\right){\mathrm{z}}_2\left(\mathrm{t}\right)}{\mathrm{aW}\left[{\mathrm{z}}_1\left(\mathrm{t}\right),{\mathrm{z}}_2\left(\mathrm{t}\right)\right]}\mathrm{dt},{\mathrm{v}}_2\left(\mathrm{t}\right)=\int \frac{\mathrm{g}\left(\mathrm{t}\right){\mathrm{z}}_1\left(\mathrm{t}\right)}{\mathrm{aW}\left[{\mathrm{z}}_1\left(\mathrm{t}\right),{\mathrm{z}}_2\left(\mathrm{t}\right)\right]}$where $\mathrm{a}=1$is a coefficient multiplying $\mathrm{z}\text{'}\text{'},\mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{lnt}+3}{\mathrm{tlnt}}$is the non-homogeneous part of the equation and $\mathrm{W}\left[{\mathrm{z}}_1\left(\mathrm{t}\right),{\mathrm{z}}_2\left(\mathrm{t}\right)\right]$is the Wronskian for functions ${\mathrm{z}}_1\left(\mathrm{t}\right)$and ${\mathrm{z}}_2\left(\mathrm{t}\right)$.

One can take ${\mathrm{C}}_1={\mathrm{C}}_2=0$.

Therefore, the particular solution is:

$$\begin{gathered} {\mathrm{z}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{z}}_1\left(\mathrm{t}\right)+{\mathrm{v}}_2\left(\mathrm{t}\right){\mathrm{z}}_2\left(\mathrm{t}\right) \\ =\left(-\frac{{\left(\mathrm{lnt}\right)}^2}{2}-3\mathrm{lnt}\right)\times \mathrm{t}+(\mathrm{lnt}+3\ln |\mathrm{lnt}\left|\right)\times \mathrm{tlnt} \\ =\frac{\mathrm{t}(\mathrm{lnt}{)}^2}{2}-3\mathrm{tlnt}(1-\ln |\mathrm{lnt}\left|\right) \end{gathered}$$

Finally, the general solution to the given differential equation is:

$\mathrm{z}\left(\mathrm{t}\right)={\mathrm{z}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{z}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{t}+{\mathrm{c}}_2\mathrm{tlnt}+\frac{\mathrm{t}(\mathrm{lnt}{)}^2}{2}-3\mathrm{tlnt}(1-\ln |\mathrm{lnt}\left|\right)$

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