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Math Help - IVP

  1. #1
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    IVP

    Hello,
    I was given the problem and to solve the initial value problem

    5x4y3dx+3x5y2dy=0y(1)=1

    I was told to find the answer in terms of F(x,y)=1 and F(x,y) does not have a constant.

    I got to the part where i've integrated both sides and got

    5ln(x) + 3ln(y)

    I'm just confused on what to do next, i tried solving for the constant which is 0

    but have no clue what the question means when its asking for it in terms of F(x,y)=1

    both answers i tried were wrong: 1- 5ln(x) + 3ln(y) & 2- 5ln(x) + 3ln(y) + 1

    thankss
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  2. #2
    Behold, the power of SARDINES!
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    Yuma, AZ, USA
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    Quote Originally Posted by washerman View Post
    Hello,
    I was given the problem and to solve the initial value problem

    5x4y3dx+3x5y2dy=0y(1)=1

    I was told to find the answer in terms of F(x,y)=1 and F(x,y) does not have a constant.

    I got to the part where i've integrated both sides and got

    5ln(x) + 3ln(y)

    I'm just confused on what to do next, i tried solving for the constant which is 0

    but have no clue what the question means when its asking for it in terms of F(x,y)=1

    both answers i tried were wrong: 1- 5ln(x) + 3ln(y) & 2- 5ln(x) + 3ln(y) + 1

    thankss
    The ODE is first order linear and separable
    5x^4y^3dx+3x^5y^2dy=0 \iff \frac{dy}{dx}=-\frac{5}{3}\cdot \frac{y}{x} \iff \frac{dy}{y}=-\frac{5}{3}\cdot \frac{dx}{x}

    Integrating both sides gives

    \ln|y|=-\frac{5}{3}\ln|x|+c \iff y=Ax^{-\frac{5}{3}}

    Using the initial condition gives A=1

    y=x^{-\frac{5}{3}}

    To put the solution into implicit form we have

    x^{-\frac{5}{3}}-y=0 \text{ or } x^{-\frac{5}{3}}-y+1=1
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