anonymous
  • anonymous
y' - 3(x^2)y = e^(x^2)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
hey got time to help with one more
anonymous
  • anonymous
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.
anonymous
  • anonymous
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).

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