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Math Help - DfEq: solution that passes through a point

  1. #1
    Member pberardi's Avatar
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    DfEq: solution that passes through a point

    Hello,
    I have this diffyQ problem that asks to find the solution that passes through a specific point. I have worked it out to where I have a ( i think correct ) equation for y, but am not sure how to deal with the constant and make sure it passes through a specific point.

    Solve: Find the solution of y' = 2(2x-y) that passes through the point P(0,1).
    Solution:
    y' = 4x -2y
    y' + 2y = 4x
    integrating factor is e^(2x)
    d/dx of [e^(2x)y] = 4xe^(2x)
    integrate both sides
    RHS by parts gives
    2xe^(2x) - e^2x
    So:
    e^(2x)y = 2xe^(2x) - e^(2x) + C
    y = 2x - 1 + Ce^(-2x)

    Could someone check this out and tell me how to get a solution that passes through that point? Thanks!
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  2. #2
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    Hello, pberardi!

    You may kick yourself when you see how easy it is . . .


    Find the solution of y' \:= \:2(2x-y) that passes through the point P(0,1).

    Your work is correct!

    Solution: . y \:= \:2x - 1 + Ce^{-2x}

    \text{"Passes through P(0,1)" means that: }\:x = 0,\:y = 1\,\text{ satisfies the equation.}

    \text{Plug them in: }\;1 \;=\;2(0) - 1 + Ce^0 \quad\Rightarrow\quad 1 \:=\:\text{-}1 + C \quad\Rightarrow\quad C \:=\:2

    Solution: . y \;=\;2x - 1 + 2e^{-2x}

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  3. #3
    Member pberardi's Avatar
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    Thank you Soroban
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  4. #4
    Senior Member
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    Hi.

    Quote Originally Posted by pberardi View Post
    Hello,
    I have this diffyQ problem that asks to find the solution that passes through a specific point. I have worked it out to where I have a ( i think correct ) equation for y, but am not sure how to deal with the constant and make sure it passes through a specific point.

    Solve: Find the solution of y' = 2(2x-y) that passes through the point P(0,1).
    Solution:
    y' = 4x -2y
    y' + 2y = 4x
    integrating factor is e^(2x)
    d/dx of [e^(2x)y] = 4xe^(2x)
    integrate both sides
    RHS by parts gives
    2xe^(2x) - e^2x
    So:
    e^(2x)y = 2xe^(2x) - e^(2x) + C
    y = 2x - 1 + Ce^(-2x)

    Could someone check this out
    Yes, it is correct.
    You just need to verify that

    y' = 2 - 2*C*e^(- 2x) = 2(2x-y)

    and tell me how to get a solution that passes through that point? Thanks!
    f(x):=y=2x - 1 + Ce^(-2x)

    P(0,1)

    f(0) = 2*0 - 1 + C *e^(-2*0) = 1

    <=> -1 + C*1 = 1

    <=> -1 + C = 1

    Find C :-)

    Regards, Rapha

    Edit: Darn, beaten by 2 minutes. What ever
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