please help find the general solution of the following first order differential equation.

dx/dt = tx + 6t(exp)-t^2

pls show working

To find the general solution of the given first-order differential equation dx/dt = tx + 6t(e^(-t^2)), we will solve it using the method of integrating factor.

Step 1: Rearrange the equation in standard form:
dx/dt - tx = 6t(e^(-t^2))

Step 2: Identify the integrating factor. In this case, the integrating factor is given by the exponential of the integral of the coefficient of x, which is t dt:
Integrating factor (I.F.) = e^(integral of -t dt)
= e^(-t^2/2)

Step 3: Multiply the original equation by the integrating factor:
e^(-t^2/2) * dx/dt - te^(-t^2/2) * x = 6te^(-t^2/2)

Step 4: Rewrite the left side of the equation as the derivative of (I.F. * x):
d/dt (e^(-t^2/2) * x) = 6te^(-t^2/2)

Step 5: Integrate both sides with respect to t:
∫ d/dt (e^(-t^2/2) * x) dt = ∫ 6te^(-t^2/2) dt

Step 6: Apply the fundamental theorem of calculus to integrate the right side:
e^(-t^2/2) * x = -3e^(-t^2/2) + C, where C is the constant of integration.

Step 7: Solve for x:
x = (-3e^(-t^2/2) + C) / e^(-t^2/2)
x = -3 + Ce^(t^2/2)

Therefore, the general solution of the given first-order differential equation is x = -3 + Ce^(t^2/2), where C is an arbitrary constant.