I am trying to solve the following differential equation using the transfer function method where we expand by partial fractions:

$$\frac{dx}{dt} + 7x = 5 cos 2t$$

The substituion is: $$(s + 7)X(s) = \frac{5s}{(s^2 + 4)}$$

Which means we have 3 variables to find:

$$\frac{A}{s + 7}$$


$$\frac{B.s + C}{s^2 + 4}$$

After checking my solution:

$A=\frac{35}{53}$, $B=\frac{50}{53}$ and $C=4$ which is wrong.

I solved it several times and kept getting the same.

The correct result is in here:

enter image description here

which gives:

enter image description here

I added images, as I don't know how to add math function on a math stack post.

The method required is the partial fraction.

  • $\begingroup$ I edited the equations to the corret formatting :) $\endgroup$ – Mateus Rocha Jul 10 '18 at 17:00
  • 3
    $\begingroup$ thank @MateusRocha. I am reading the edit so I can learn how we write them for the next time. $\endgroup$ – alim1990 Jul 10 '18 at 17:01
  • $\begingroup$ Here's a pretty thorough tutorial on MathJax formatting; enjoy! math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Robert Howard Jul 10 '18 at 17:03
  • $\begingroup$ the book's answer is correct wolframalpha.com/input/… $\endgroup$ – Isham Jul 10 '18 at 17:09

$$(s + 7)X(s) = \frac{5s}{(s^2 + 4)}\implies X(s)=\frac{5s}{(s+7)(s^2 + 4)}$$

$$\frac{5s}{(s+7)(s^2 + 4)}=\frac{A}{(s+7)} +\frac{Bs+C}{(s^2 + 4)}$$

To find $A$ cover $(s+7)$ and let $s=-7$ in $$\frac{5s}{(s+7)(s^2 + 4)}$$ to get $$A=\frac {-35}{53}$$ ( That is where you had it wrong )

Now proceed to find $B$ and $C$ by assigning other values of $s$.

  • $\begingroup$ That's it. Thanks for the explanation. $\endgroup$ – alim1990 Jul 10 '18 at 17:23
  • $\begingroup$ @droidnation Thanks for your attention. $\endgroup$ – Mohammad Riazi-Kermani Jul 10 '18 at 17:29

If we write it as: $$\frac{5s}{(s+7)(s^2+4)}=\frac{A}{s+7}+\frac{Bs+C}{s^2+4}$$ We can multiply both sides by $(s+7)(s^2+4)$: $$5s=A(s^2+4)+(Bs+C)(s+7)$$ $$5s=As^2+4A+Bs^2+7Bs+Cs+7C$$ $$0s^2+5s+0=(A+B)s^2+(7B+C)s+7C+4A$$ So we have that $A+B=0$ and $7B+C=5$ and $7C+4A=0$. From the first one, we have that $A=-B$, so we have that $7B+C=5$ and $7C-4B=0$. And I think you can solve this system of 2 linear equations.

  • 1
    $\begingroup$ Oh. My error was by multipliying with one factor each time and not with $(s+7)(s^2+4)$ $\endgroup$ – alim1990 Jul 10 '18 at 17:02

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