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Thread: Want help in solving this Diferrential equation by variable separable method!

  1. #1
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    Red face Want help in solving this Diferrential equation by variable separable method!

    How can I solve this euqation by variable separable ? I know after separating variable it will be solved by partial fraction by I stuck at the values of A and B on the side of y!


    dy/dx = (y3 + 2y) / (x2 ​+ 3x )
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    Re: Want help in solving this Diferrential equation by variable separable method!

    $\dfrac{1}{y^3+2y}=\dfrac{A}{y}+\dfrac{By+C}{y^2+2 } $

    The numerator is always a polynomial of degree no greater than one less than the degree of the denominator.

    $A=\dfrac{1}{2},B=-\dfrac{1}{2},C=0$
    Last edited by SlipEternal; Nov 8th 2017 at 03:26 AM.
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    Re: Want help in solving this Diferrential equation by variable separable method!

    May i know the steps of solution ?
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    Re: Want help in solving this Diferrential equation by variable separable method!

    Quote Originally Posted by damna97 View Post
    May i know the steps of solution ?
    It is one step.
    $A (y^2+2)+(By+C)y=1$

    $2A=1$ (coefficient of $y^0$)
    $C=0$ (coefficient of $y^1$)
    $A+B=0$ (coefficient of $y^2$)
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  5. #5
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    Re: Want help in solving this Diferrential equation by variable separable method!

    Step 1: take a Calculus class and learn to integrate!

    You say yourself this equation is "separable"- so separate it. \frac{dy}{dx}= \frac{y^3+ 2y}{x^2+ 3x} becomes \frac{dy}{y^3+ 2y}= \frac{dx}{x^2+ 3x}.

    \frac{dy}{y(y^2+ 2)}= \frac{dx}{x(x+ 3)}

    Now use "partial fractions" as you learned in Calculus to integrate those. That is what SlipEternal did.
    \frac{1}{y(y^2+ 2)}= \frac{A}{y}+ \frac{By+ C}{y^2+ 2}
    Multiply both sides by y(y^2+ 2) to get 1= A(y^2+ 2)+ (By+ C)y. You need three equations to determine the three unknowns, A, B, and C, so substitute three different values for y to get them. If y= 0, 1= 2A. If y= 1, 1= 3A+ B+ C. If y= -1, 1= 3A+ B- C.

    From 1= 2A, A= 1/2. The other two equations become 1= (3/2)+ B+ C or B+ C= 1- 3/2= -1/2 and 1= (3/2)+ B- C or B- C= 1- 3/2= -1/2. Adding those two equations 2B= -1 so B= -1/2. Subtracting 2C= 0 so C= 0 as slip eternal said.
    \frac{1}{y(y^2+ 2)}= \frac{1/2}{y}- \frac{1/2}{y^2+ 2}.

    On the other side you have \frac{1}{x(x+ 3)}= \frac{A}{x}+ \frac{B}{x+ 3}. (Of course, these are different "A" and "B" than before.)
    Multiply on both sides by x(x+ 3) to get 1= A(x+ 3)+ Bx. If x= 0, that becomes 1= 3A so A= \frac{1}{3}. If x= -3, that becomes 1= -3B so B= -\frac{1}{3}
    \frac{1}{x(x+ 3)}= \frac{1/3}{x}- \frac{1/3}{x+ 3}.

    So your equation finally reduces to
    \frac{1}{2}\frac{dy}{y}- \frac{1}{2}\frac{dy}{y^2+ 2}= \frac{1}{3}\frac{dx}{x}- \frac{1}{3}\frac{dx}{x+ 3}.

    Can you integrate that?
    Thanks from damna97
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    Re: Want help in solving this Diferrential equation by variable separable method!

    Thankyou! Now I can solve this easily ! ☺️
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    Re: Want help in solving this Diferrential equation by variable separable method!

    Quote Originally Posted by HallsofIvy View Post
    Step 1: take a Calculus class and learn to integrate!

    You say yourself this equation is "separable"- so separate it. \frac{dy}{dx}= \frac{y^3+ 2y}{x^2+ 3x} becomes \frac{dy}{y^3+ 2y}= \frac{dx}{x^2+ 3x}.

    \frac{dy}{y(y^2+ 2)}= \frac{dx}{x(x+ 3)}

    Now use "partial fractions" as you learned in Calculus to integrate those. That is what SlipEternal did.
    \frac{1}{y(y^2+ 2)}= \frac{A}{y}+ \frac{By+ C}{y^2+ 2}
    Multiply both sides by y(y^2+ 2) to get 1= A(y^2+ 2)+ (By+ C)y. You need three equations to determine the three unknowns, A, B, and C, so substitute three different values for y to get them. If y= 0, 1= 2A. If y= 1, 1= 3A+ B+ C. If y= -1, 1= 3A+ B- C.

    From 1= 2A, A= 1/2. The other two equations become 1= (3/2)+ B+ C or B+ C= 1- 3/2= -1/2 and 1= (3/2)+ B- C or B- C= 1- 3/2= -1/2. Adding those two equations 2B= -1 so B= -1/2. Subtracting 2C= 0 so C= 0 as slip eternal said.
    \frac{1}{y(y^2+ 2)}= \frac{1/2}{y}- \frac{1/2}{y^2+ 2}.

    On the other side you have \frac{1}{x(x+ 3)}= \frac{A}{x}+ \frac{B}{x+ 3}. (Of course, these are different "A" and "B" than before.)
    Multiply on both sides by x(x+ 3) to get 1= A(x+ 3)+ Bx. If x= 0, that becomes 1= 3A so A= \frac{1}{3}. If x= -3, that becomes 1= -3B so B= -\frac{1}{3}
    \frac{1}{x(x+ 3)}= \frac{1/3}{x}- \frac{1/3}{x+ 3}.

    So your equation finally reduces to
    \frac{1}{2}\frac{dy}{y}- \frac{1}{2}\frac{dy}{y^2+ 2}= \frac{1}{3}\frac{dx}{x}- \frac{1}{3}\frac{dx}{x+ 3}.

    Can you integrate that?
    Slight typo. Halls wrote that $B = -\dfrac{1}{2}, C=0$, then assigned $B=0, C=-\dfrac{1}{2}$. It should be:

    \dfrac{1}{y(y^2+2)} = \dfrac{1/2}{y} - \dfrac{y/2}{y^2+2}

    .
    .
    .
    \dfrac{1}{2}\left( \dfrac{dy}{y} - \dfrac{y}{y^2+2}\right) = \dfrac{1}{3}\left(\dfrac{dx}{x} - \dfrac{dx}{x+3} \right)

    So, the LHS is actually easier than what Halls wrote (it will not require an inverse trigonometric integration).
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    Re: Want help in solving this Diferrential equation by variable separable method!

    Thanks.
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