The ODE is separable:
[tex]\dfrac{\mathrm dy}{\mathrm dx}=(x-2)(y-10)\implies\dfrac{\mathrm dy}{y-10}=(x-2)\,\mathrm dx[/tex]
Integrating both sides gives
[tex]\ln|y-10|=\dfrac{(x-2)^2}2+C[/tex]
[tex]\implies y-10=e^{(x-2)^2/2+C}=Ce^{(x-2)^2/2}[/tex]
[tex]\implies y=10+Ce^{(x-2)^2/2}[/tex]
Given that [tex]y(0)=5[/tex], we find
[tex]5=10+Ce^{(-2)^2/2}\implies-5=Ce^2\implies C=-5e^{-2}[/tex]
so that
[tex]\boxed{y(x)=10-5e^{(x-2)^2/2-2}}[/tex]
