Question

y' - 3(x^2)y = e^(x^2)

Answer 1

hey got time to help with one more

Answer 2

Is it e^(x^2) or e^(x^3)? For e^(x^2) I don't think it's solvable by elementary methods. For e^(x^3), it is.

Answer 3

This is called the method of integrating factors.

The idea is that we'd like to rewrite the Left Hand Side of
y' +f(x) y = g(x)

as the derivative of a product of y with some other function h(x).
Notice that the derivative of y(x)h(x) is y’h + h’y.

This kind of looks like y’ + f(x)*y.

But to make that look even more like y’h + h’y,
we can multiply y’ + f(x)*y by the function h (to be determined)
to get h[y’ + f(x)*y] = y’h + (h*f)*y.

So what should our function h be ?
Our h should satisfy (hf) = h’ to match the expressions y’h + (hf)*y and y’h + h’y.

Solving hf = h’ is easy since its separable.

In this case, if we have y’ – 3(x^2)y = e^(x^3).
So we’d like to multiply both sides of the equation by some h(x).
And that h(x) should satisfy h*(-3x^2) = h’.
Solving this gives h(x) = e^(-x^3).

Then we see that multiplying both sides of our equation by h(x) gives
e^(-x^3) * y’ + [-3(x^2)* e^(-x^3)] * y = e^(0) = 1

And, just as we wanted, the Left Hand Side is
just the derivative of the product [e^(-x^3)*y].

If we integrate both sides with respect to x,
we have that e^(-x^3)*y = x + C,
so multiplying both sides by e^(x^3), we see that y = xe^(x^3) + Ce^(x^3).