My best guess at interpreting this question is that you need to solve
[tex]x^2y''=2xy'-y[/tex]
which seems like a reasonable interpretation as this ODE is of a well-known type (Cauchy-Euler). And it looks like you're given two initial conditions, [tex]y(1)=4[/tex] and [tex]y'(1)=2[/tex].
[tex]x^2y''-2xy'+y=0[/tex]
Substitute [tex]y=x^m[/tex], so that [tex]y'=mx^{m-1}[/tex] and [tex]y''=m(m-1)x^{m-2}[/tex]. Then plugging these into the ODE gives
[tex]m(m-1)x^m-2mx^m+x^m=0[/tex]
[tex]m(m-1)-2m+1=0[/tex]
[tex]m^2-3m+1=0[/tex]
[tex]\implies m_1=\dfrac{3+\sqrt5}2,m_2=\dfrac{3-\sqrt5}2[/tex]
So the homogeneous ODE has general solution
[tex]y=C_1e^{m_1x}+C_2e^{m_2x}[/tex]
This solution has derivative
[tex]y'=m_1C_1e^{m_1x}+m_2C_2e^{m_2x}[/tex]
From the initial conditions, we get
[tex]4=C_1e^{m_1}+C_2e^{m_2}[/tex]
[tex]2=m_1C_1e^{m_1}+m_2C_2e^{m_2}[/tex]
Solving this system gives
[tex]C_1=\dfrac{2e^{-m_1}(1-2m_2)}{m_1-m_2},C_2=\dfrac{2e^{-m_2}(2m_1-1)}{m_1-m_2}[/tex]
thus giving the particular solution,
[tex]\boxed{y(x)=\dfrac{2(1-2m_2)}{m_1-m_2}e^{m_1(x-1)}+\dfrac{2(2m_1-1)}{m_1-m_2}e^{m_2(x-1)}}[/tex]
