How do i work this differential equation

t dx/dt +2x(t) = 4e^t^2

Solve this differential equation by using an integrating factor:
t(dx / dt) + 2x = 4℮^t²
dx / dt + 2x / t = 4℮^t² / t
dx / dt + P(t)x = f(t)
P(t) = 2 / t
f(t) = 4℮^t² / t
I(t) = ℮^[∫ P(t) dt]
I(t) = ℮^(∫ 2 / t dt)
I(t) = ℮^(2ln|t|)
I(t) = ℮^(lnt²)
I(t) = t²
I(t)x = ∫ I(t)f(t) dt
t²x = ∫ 4t℮^t² dt

Integrate the expression on the right by substitution:
∫ 4t℮^t² dt = 2 ∫ 2t℮^t² dt
Let u = t²,
du / dt = 2t
du = 2t dt
∫ 4t℮^t² dt = 2 ∫ ℮ᵘ du
∫ 4t℮^t² dt = 2℮ᵘ + C
Since u = t²,
∫ 4t℮^t² dt = 2℮^t² + C

Find the general solution using the above result:
t²x = ∫ 4t℮^t² dt
t²x = 2℮^t² + C
t²x = C + 2℮^t²
x = (C + 2℮^t²) / t²

You have:

t*dx/dt + 2x = 4*exp(t^2)

dx/dt + (2/t)*x = (4/t)*exp(t^2)

This is a first-order, linear ODE that we can try to solve using an integrating factor. For an equation of the form:

dy/dx + a(x)*y = b(x)

the integrating factor is given by:

p(x) = exp(INTEGRAL of {a(x) dx})

and the solution is given by:

y(x) = (1/p(x))*INTEGRAL of {p(x)*b(x) dx}

In this case,

p(t) = exp(INTEGRAL of {2/t dt}) = exp(2ln(t)) = exp(ln(t^2)) = t^2

The solution is then given by:

x(t) = (1/t^2)*INTEGRAL of {(4*t*exp(t^2) dt}

To do the integral, let u = t^2, then du = 2t dt

x(t) = (1/t^2)*[2*exp(t^2) - c]

where c is the constant of integration.