<s>[INST] $y(x)$ is a solution of the differential equation Given that $y(x)$ is a solution of the differential equation
$$
x^2 y^{\prime \prime}+x y^{\prime}-4 y=x^2
$$
on the interval $(0, \infty)$ such that $\lim _{x \rightarrow 0^{+}} y(x)$ exists and $y(1)=1$. I need to find the value of $y^{\prime}(1)$.
$$
\begin{aligned}
& x^2 y^{\prime \prime}+x y^{\prime}-4 y=x^2 \\
& \left(x^2 D^2+x D-4\right) y=x^2 \\
& z=\log x \quad \begin{aligned}
x^2 D^2 & =D^{\prime}\left(D^{\prime}-1\right) \\
& =D^{\prime 2}-D^{\prime}
\end{aligned} \\
& \left(D^{\prime 2}-D^{\prime}+D^{\prime}-4\right) y=\left(e^z\right)^2 \\
& \left(D^{\prime 2}-4\right) y=e^{2 z} \\
& \text { Complementary Function } \\
&  \\
& m^2-4=0 \\
& m^2=4 \\
& m=\pm 2 \\
& \text { c. } F=c_1 e^{2 z}+c_2 e^{-2 z} \\
& \begin{array}{l}
=c_1\left(e^z\right)^2+c_2\left(e^z\right)^{-2} \\
=c_1 x^2+c_2 x^{-2}
\end{array} \mid \\
& =c_1 x^2+\frac{1}{x^2} c_2 \\
&
\end{aligned}
$$
$$
\begin{aligned}
P \cdot I & =\frac{1}{D^{\prime 2}-4} e^{2 z} \\
& =\frac{1}{(2)^2-4} e^{2 z}  \\
& =\frac{1}{4-4} e^{2 z} \quad \text {  } \\
& =\frac{z}{2 D^{\prime}} e^{2 z} \\
& =\frac{z}{4} e^{2 z} \\
P I & =\frac{\log x}{4} \cdot x^2
\end{aligned}
$$
General soln,
$$
\begin{aligned}
y(x) & =C \cdot F+P \cdot I \\
& =c_1 x^2+c_2 \frac{1}{x^2}+\frac{\log x}{4} x^2
\end{aligned}
$$
Further how to get the solution ?
Thanks in advance.
 [/INST] You know that $\displaystyle\lim_{x\to 0^+}y(x)$ exists.
It is equivalent to say that, $\displaystyle\lim_{x\to 0^+}\left(c_1x^2+c_2\frac{1}{x^2}+\frac{\ln(x)}{4}x^2\right)$ exists.
You know that $\displaystyle\lim_{x\to 0^+}\frac{\ln(x)}{4}x^2=0$.
Now, you know that $\displaystyle\lim_{x\to 0^+}\left(c_1x^2+c_2\frac{1}{x^2}\right)$ exists.
The only way is that $c_2=0$
This gives: $y(x)=c_1x^2+\frac{\ln(x)}{4}x^2$
By evaluating in $x=1$, you get $y(1)=c_1$. However, $y(1)=1$. So $c_1=1$.
Finally, $y(x)=x^2+\frac{\ln(x)}{4}x^2$
I think you will be able to derive and find $y^\prime(1)$ by yourself.
</s>