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Math Help - differential equations

  1. #1
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    differential equations

    hey all!

    ive got two differentail equations which i really cannot do, but my friend suggested using this website so here goes!

    a) xy(dy/dx) - 2(y+3)=0

    b) (dy/dx) - y=xy5 (i think its a special one but i dont know how to do it!)

    sorry about the writing style, i dont know how to use the special maths tool yet but i hope you understand what i mean :P

    thank you
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  2. #2
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    Quote Originally Posted by emily28 View Post
    a) [I]xy(dy/dx) - 2(y+3)=0
    xyy' = 2(y+3)
    \frac{yy'}{2(y+3)} = \frac{1}{x} (or y=-3 is solution).
    THis is seperable.
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  3. #3
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    Quote Originally Posted by emily28 View Post
    hey all!

    b) (dy/dx) - y=xy5
     <br />
\begin{aligned}<br />
y' - y &= 5xy\\<br />
y' &= y(5x + 1)\\<br />
\frac{1}<br />
{y}~dy &= (5x + 1)~dx<br />
\end{aligned}<br />

    Now it's easier.
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  4. #4
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    y'-y=xy^5\iff\frac{y'}{y^5}-\frac1{y^4}=x

    If this is the equation, then set u=\frac1{y^4}
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  5. #5
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    oh right, but im still unsure where to go from here becuase its seperable but im still slightly confused
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  6. #6
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    But you didn't tell me what's the right equation
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  7. #7
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    oh right sorry it was

    y'-y=xy to the power 5

    does this help!!
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  8. #8
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    Quote Originally Posted by Krizalid View Post
    y'-y=xy^5\iff\frac{y'}{y^5}-\frac1{y^4}=x

    If this is the equation, then set u=\frac1{y^4}
    Follow this, 'cause it's a Bernoulli's differential equation.
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  9. #9
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    i understand it upto this point and know its bernolli equation, but i dont know how to solve them at all! could anyone suggest where i should go next
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  10. #10
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    Quote Originally Posted by emily28 View Post
    oh right sorry it was

    y'-y=xy to the power 5

    does this help!!
    Quote Originally Posted by Krizalid View Post
    Follow this, 'cause it's a Bernoulli's differential equation.
    Quote Originally Posted by emily28 View Post
    i understand it upto this point and know its bernolli equation, but i dont know how to solve them at all! could anyone suggest where i should go next
    Let u = y^{-4} (ie. y = u^{-1/4}.)

    Then \frac{du}{dx} = -4 \cdot y^{-5} \frac{dy}{dx}

    Turning this around we get:
    \frac{dy}{dx} = - \frac{1}{4}y^5 \frac{du}{dx} = - \frac{1}{4} u^{-5/4} \frac{du}{dx}

    So the differential equation becomes:
     y'-y=xy^5

    - \frac{1}{4} u^{-5/4} \frac{du}{dx} - <br />
u^{-1/4} = x \cdot u^{-5/4}<-- Multiply both sides by u^{5/4}.

    - \frac{1}{4} \frac{du}{dx} - <br />
u = x

    \frac{du}{dx} + 4 u = -4x

    Now solve this for u(x) and when you get that find y(x).

    -Dan
    Last edited by topsquark; July 11th 2007 at 05:10 AM.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post

    \frac{du}{dx} + 4 u = -4x

    Now solve this for u(x) and when you get that find y(x).

    -Dan
    i did the manipulation slightly different from what you did, but we ended up with the same thing. i used the modified equation that Krizalid developed to plug the substitutions into. i think it works out nicer that way, but what do i know, maybe it's just nicer in my weird mind
    Last edited by Jhevon; July 11th 2007 at 05:19 AM.
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  12. #12
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    hey, sorry guys im still slightly stumped in how to do the next part...
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  13. #13
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by emily28 View Post
    hey, sorry guys im still slightly stumped in how to do the next part...
    it's a first order differential equation. multiply through by the integrating factor and simplify. (You know how to find integrating factors right? ...do you know what an integrating factor is?)
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  14. #14
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    \frac{du}{dx} + 4 u = -4x
    Quote Originally Posted by emily28 View Post
    hey, sorry guys im still slightly stumped in how to do the next part...
    Solve the homogeneous equation first:
    \frac{du_h}{dx} + 4 u_h = 0

    \frac{du_h}{dx} = -4 u_h

    What function could u_h(x) be? Since we know that \frac{d}{dx}e^{bx} = be^{x} the function u_h(x) must be an exponential. Thus
    u_h(x) = Ae^{bx}

    Plugging this into the homogeneous equation gives:
    Abe^{bx} = -4Ae^{bx}

    This leaves us with b = -4. So u_h(x) = Ae^{-4x}.

    Now for the particular solution of \frac{du}{dx} + 4 u = -4x. The RHS is a polynomial so we would guess that the particular solution u_p(x) will be a polynomial. So try u_p(x) = cx + d. Plugging this into the differential equation gives:
    c + 4 (cx + d) = -4x

    4cx + (c + 4d) = -4x

    So we have the simultaneous equations
    4c = -4
    and
    c + 4d = 0

    Thus c = -1 and d = 1/4. Thus u_p(x) = -x + \frac{1}{4}.

    Since the differential equation is linear the most general solution to the differential equation will be the sum of the homogenous and particular solutions. Thus
    u(x) = u_h(x) + u_p(x) = Ae^{-4x} - x + \frac{1}{4}

    Now, to get y(x) we know that y(x) = -\frac{1}{u^{1/4}}. Thus we finally have that
    y(x) = -\frac{1}{(e^{-4x} - x + \frac{1}{4})^{1/4}} = -\frac{1}{\sqrt[4]{e^{-4x} - x + \frac{1}{4}}}

    -Dan
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  15. #15
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    i dont that i was meant to look at particular solutions and expenituals, is there another way to do it using perhaps a different method, as i havent covered this in class and dont think the teacher would want me to approach it this way?
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