Multiply both sides of the ODE
[tex]x'+2tx+5t[/tex]
by [tex]e^{t^2}[/tex]:
[tex]e^{t^2}x'+2te^{t^2}x=5te^{t^2}[/tex]
Now the left side can be condensed as the derivative of a product:
[tex]\left(e^{t^2}x\right)'=5te^{t^2}[/tex]
Integrate both sides, then solve for x :
[tex]e^{t^2}x=\dfrac52e^{t^2}+C[/tex]
[tex]\implies x(t)=\dfrac52+Ce^{-t^2}[/tex]
Given that x(0) = 8, we find
[tex]8=\dfrac52+Ce^0\implies C=\dfrac{11}2[/tex]
so that the particular solution to this IVP is
[tex]\boxed{x(t)=\dfrac{5+11e^{-t^2}}2}[/tex]
