Question

Solve the initial-value problem. x' + 2tx = 5t, x(0) = 8 x(t) =

Answer 1

Multiply both sides of the ODE

$x'+2tx+5t$

by $e^{t^2}$:

$e^{t^2}x'+2te^{t^2}x=5te^{t^2}$

Now the left side can be condensed as the derivative of a product:

$\left(e^{t^2}x\right)'=5te^{t^2}$

Integrate both sides, then solve for x :

$e^{t^2}x=\dfrac52e^{t^2}+C$

$\implies x(t)=\dfrac52+Ce^{-t^2}$

Given that x(0) = 8, we find

$8=\dfrac52+Ce^0\implies C=\dfrac{11}2$

so that the particular solution to this IVP is

$\boxed{x(t)=\dfrac{5+11e^{-t^2}}2}$