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Math Help - Initial Value Problem

  1. #1
    Len
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    Initial Value Problem

    I missed the lecture and the text book goes over this very quickly...

    \frac{dy}{dt}=\frac{1}{2y+3},  y(0)=1

    This is an autonomous and therefore separable equation so

    dt=(2y+3)dy

    \int dt= \int (2y+3) dy

    t=y^2+3y+C

    If I have done this right, what is the next step? If not please correct :$

    Thank You
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Len View Post
    I missed the lecture and the text book goes over this very quickly...

    \frac{dy}{dt}=\frac{1}{2y+3},  y(0)=1

    This is an autonomous and therefore separable equation so

    dt=(2y+3)dy

    \int dt= \int (2y+3) dy

    t=y^2+3y+C

    If I have done this right, what is the next step? If not please correct :$

    Thank You
    It all is correct so far. Just keep in mind that y is the subject of the equation. So we can rewrite what you have as y^2+3y=t+C. Note that if you complete the square, you end up with \left(y+\tfrac{3}{2}\right)^2=t+C-\tfrac{3}{2}\implies \left(y+\tfrac{3}{2}\right)^2=t+K, where K is another constant.

    Now, applying the initial condition y(0)=1, we have \tfrac{25}{4}=K

    Therefore, \left(y+\tfrac{3}{2}\right)^2=t+\tfrac{25}{4}

    Does this make sense?
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  3. #3
    Len
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    Quote Originally Posted by Chris L T521 View Post
    It all is correct so far. Just keep in mind that y is the subject of the equation. So we can rewrite what you have as y^2+3y=t+C. Note that if you complete the square, you end up with \left(y+\tfrac{3}{2}\right)^2=t+C-\tfrac{3}{2}\implies \left(y+\tfrac{3}{2}\right)^2=t+K, where K is another constant.

    Now, applying the initial condition y(0)=1, we have \tfrac{25}{4}=K

    Therefore, \left(y+\tfrac{3}{2}\right)^2=t+\tfrac{25}{4}

    Does this make sense?
    I think so, that can be rewrote as

    y =\sqrt{t+k}-\frac{3}{2}
    then using initial condition
    (\frac{5}{2})^2=k

    Awesome, thank you very much.

    Although looking back I didn't need to do that, simply substituting y=1 and t=0 would give me the answer
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  4. #4
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    Initial Value Problem

    Hello Len
    Quote Originally Posted by Len View Post
    I missed the lecture and the text book goes over this very quickly...

    \frac{dy}{dt}=\frac{1}{2y+3},  y(0)=1

    This is an autonomous and therefore separable equation so

    dt=(2y+3)dy

    \int dt= \int (2y+3) dy

    t=y^2+3y+C

    If I have done this right, what is the next step? If not please correct :$

    Thank You
    Perfect, so far!

    y(0) means the value of y when t = 0. So you can just plug this into your equation:

    When t = 0, y = 1: 0 = 1^2 + 3\times 1 + C.

    So C = ...?

    Put this value of C back into your equation, and there's your answer for t in terms of y.

    Grandad
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