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Math Help - Calculus Integration problem. NO clue

  1. #1
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    Calculus Integration problem. NO clue

    Question is as follows :

    Let f be the function satisfying f ' (x) = -3xf(x) for all real numbers x, with f(1) = 4 and lim as x -> infinity of f(x) = 0.

    a.) Evaluate the integral of -3xf(x) dx with bounds 1 and infinity.

    b.) Use Euler's method, starting at x = 1 with a step size of .5 to approximate f(2)

    c.) Write an expression for y = f(x) by solving the differential equation dy/dx = -3xy with the initial condition f(1) = 4


    I dont really need help with b, once I figure out the equation I can do that one. But A and C have me completely stumped.
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  2. #2
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    Quote Originally Posted by Naveed103 View Post
    Question is as follows :

    Let f be the function satisfying f ' (x) = -3xf(x) for all real numbers x, with f(1) = 4 and lim as x -> infinity of f(x) = 0.

    a.) Evaluate the integral of -3xf(x) dx with bounds 1 and infinity.

    b.) Use Euler's method, starting at x = 1 with a step size of .5 to approximate f(2)

    c.) Write an expression for y = f(x) by solving the differential equation dy/dx = -3xy with the initial condition f(1) = 4


    I dont really need help with b, once I figure out the equation I can do that one. But A and C have me completely stumped.
    a.) Integrate both sides of f ' (x) = -3xf(x): f(x) = \int -3x f(x) \, dx.

    Therefore \int_{1}^{\infty} -3x f(x) \, dx = \lim_{a \rightarrow +\infty} f(a) - f(1) = 0 - 4 = -4.

    c.) dy/dx = -3xy is seperable: \frac{1}{y}\, dy = -3x \, dx etc. etc.
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  3. #3
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    thanks a lot for your help on a. So if I integrate c, I get y/4 = -3/2 x^2 + c. f(1) = 4, and c = 10, so the final solution is y/4 = -3/2 x^2 + 5/2 or y = -6x^2 + 10?
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  4. #4
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    Quote Originally Posted by Naveed103 View Post
    thanks a lot for your help on a. So if I integrate c, I get y/4 = -3/2 x^2 + c. f(1) = 4, and c = 10, so the final solution is y/4 = -3/2 x^2 + 5/2 or y = -6x^2 + 10?
    First of all, you can check a solution by substituting it back into the DE. Does your answer work?

    I cannot see how you could possibly figure that the integral of 1/y is y/4, which I guess preempts the answer to my question above .......
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  5. #5
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    oh I'm sorry. The 1/y I wrote on my paper looks like 1/4, ><. I've got bad handwriting and it's late. so if you integrate 1/y you get lny = -3/2 x^2 + c. If I solve for c I get 2.88, which seems like a really odd number. I'm sorry if these questions are stupid, I'm not very good at this stuff.
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