Can someone please help!!! How do you solve for the inverse laplace of 1/(s(s+4)(s+4))
@Spacelimbus
\[\Large \frac{1}{s(s+4)^2} \]?
yes
Partial fraction decomposition \[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2}\]
so you'd get A(S+4)(S+4)^2 +B(s)(S+4)^2 +C(s)(S+4)^2
If you multiply through you are left with \[\Large 1=A(s+4)^2+B(s+4)+Cs \]
I thought you are supposed to multiply A by the values under B and C?
I made a mistake above, but no you want to solve for the coefficients so you might get an easier expression
\[ \Large 1=A(s+4)^2+Bs(s+4)+Cs\]
better now it seems
wouldnt you get A=0, B=0 and C=-1/4
Sorry for the mistake again, I messed up the above form, so if you solved for that your answer will be wrong.
\[\Large 1=(A+B)s^2 +(8A+4B+C)s+16A \]
could you please explain how you go that
\[A+B=0 \\ 8A+4B+C=0 \\ 16A=0 \]
Yes, I multiplied the above form by s(s+4)^2 (make sure you cancel on the right hand side right) and then you factor likewise terms. When you did that, you can set both sides of the equation equal, they are equal if and only if their coefficients match.
\[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2} \] General setup of partial fraction decomposition, multiply through by \(s(s+4)^2\), it will disappear entirely on the left hand side of the equation and piecewise on the righthandside of the equation.
Then expand the righthand side and factor likewise terms, set the coefficients equal to the coefficients on the lefthand side to make a valid equation out of it LHS=RHS, in this case a lot of the coefficients have to be zero, because there is only 1 on the LHS.
you will end up with a 3x3 system of linear equations. \[A+B=0 \\ 8A+4B+C=0 \\ 16A=1 \] *corrected the last line*
okay so far?
sorry im still confused on how you got the equation
can you please show me by the work?
The idea of partial fraction decomposition is that you find a different way of writing the quotient, so in the end there will be an easier form for it which is especially useful in integral calculus and for LaPlace Transformations. The idea is that you completely factor the denominator and then you split it up, so each new quotient becomes its own denominator
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