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Thread: Partial fractions question

  1. #1
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    Partial fractions question

    Hello, I am in a differential equations class need to find the inverse Laplace of
    $\displaystyle \frac{1}{s(s^2+5)}$, but have forgotten how to do partial fractions with quadratic factors. What I got so far:

    $\displaystyle \frac{1}{s(s^2+5)}=\frac{A}{s}+\frac{Bs+C}{s^2+5}$

    $\displaystyle A(s^2+5)+(Bs+C)s=1$

    Letting s=0 I get $\displaystyle A=\frac{1}{5}$, but where do I go from there?
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  2. #2
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    Lightbulb Re: Partial fractions question

    $\displaystyle \frac{1}{s\left(s^2+5\right)} = \frac{1}{5 s}-\frac{s}{5 \left(5+s^2\right)}$
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    Re: Partial fractions question

    Thanks! Could you explain how you got that?
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    Re: Partial fractions question

    Equate coefficients of the various powers of $\displaystyle s$ across the identity.

    The number of $\displaystyle s^{2}$ on the LHS have to equal the number of $\displaystyle s^{2}$ on the RHS.

    That gives you $\displaystyle A+B=0.$

    Then equate coefficients of $\displaystyle s$ to get the value of $\displaystyle C.$

    Alternatively you could substitute two other values for $\displaystyle s$ and solve the resulting simultaneous equations.
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  5. #5
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    Re: Partial fractions question

    You have $\displaystyle A(s^2+ 5)+ (Bs+ C)s= 1$ which is to be true for all s. So take s to be some simple numbers.

    If s= 0, then $\displaystyle A(0+ 5)+ (B(0)+ C)0= 5A= 1$. Putting that value into the equation, $\displaystyle (1/5)(s^2+ 5)+ (Bs+ C)s= (1/5+ B)s^2+ Cs+ 1= 1$ or $\displaystyle (1/5+ B)s^2+ Cs= 0$.
    Taking s= 1, that becomes 1/5+ B+ C= 0 or B+ C= -1/5. Taking s= -1, it is 1/5+ B- C= 0 or B- C= 1/5. Adding those, 2B= 0 so B= 0. Subtracting, 2C= -2/5 so C= -1/5.

    Another way to get that is to multiply out $\displaystyle A(s^2+ 5)+ (Bs+ C)s=(A+ B)s^2+ Cs+ 5A= 0s^2+ 0s+ 1$. Now, because that is true for all s, we must have "corresponding coefficients" equal: A+ B= 0, C= 0, 5A= 1. That gives A= 1/5 and then (1/5)+ B= 0 so B= -1/5.
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    Re: Partial fractions question

    Aah! Thanks a lot!
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