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.
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.