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The differential equation is as follows-

$$\frac{d^2 x}{dt^2} + 5 \frac{dx}{dt} + 6x = e^t $$

I use laplace transform to make it to become - $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$ where $X(s)$ is the Laplace transform of $X(t)$

So now I am trying to find $X(t)$ using inverse transform.

From partial fractions-

$X(s) = \frac{1}{(s-1)(s+3)(s+2)} = \frac{A}{s-1} + \frac{B}{s+3} + \frac{C}{s+2} $

Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $

I am stuck from here on how to carry on this partial fraction

Can I sub all s values to be 0 ?

For example

$1 = A(0+3)(0+2)$

$1= B(0-1)(0+2) $

$1 = C (0-1)(0+3) $

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    $\begingroup$ You can expand everything, and compare the coefficients of $s^2$,$s$,$1$ on both sides. Or you can even substitute some values for $s$. $\endgroup$
    – Botond
    Jul 22, 2018 at 21:02

3 Answers 3

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$ s=0$ is not the best choice

There are three values to assign to $s$ which makes our life very easy.

$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$

$$ s=1 \implies 12A=1 \implies A=\frac {1}{12}$$

$$ s=-3 \implies 4B=1 \implies B=\frac {1}{4}$$

$$ s=-2 \implies -3C=1 \implies C=\frac {-1}{3}$$

Now you proceed with the inverse Laplace Transform.

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  • $\begingroup$ For example , why can’t I use -4 to find B ? Apparently I will get B= 1/10 which is different if I use -3. is this ok? $\endgroup$
    – user185692
    Jul 22, 2018 at 21:34
  • $\begingroup$ Or must I only identify -3 and use -3 ? $\endgroup$
    – user185692
    Jul 22, 2018 at 21:35
  • $\begingroup$ @user185692 If you use -4 then you will have $1=A(-4+3)(-4+2)+B(-4−1)(-4+2)+C(-4−1)(-4+3)$ which does not give you a simple solution for $B$. As you see $ A$ and $C$ are still present. $\endgroup$
    – Mohammad Riazi-Kermani
    Jul 22, 2018 at 21:39
  • $\begingroup$ Why must I sub all the s values to be -4 whereas you sub 1,-3,-2 for A,B, and C respectively ? I was asking must these 3 numbers be fixed to be 1,-3,-2 ? Like for example, could I use 1,-4,-2 respectively for A, B, C. Thanks!!! $\endgroup$
    – user185692
    Jul 22, 2018 at 21:48
  • $\begingroup$ @user185692 When you let $s=1$ the coefficients of $B$ and $C$ become $0$. The only survivor is $A$. $\endgroup$
    – Mohammad Riazi-Kermani
    Jul 22, 2018 at 21:51
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So, if we have that $$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$ We can expand everything to get $$1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$$ $$1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$$ So we must have that $A+B+C=0$, $5A+B+2C=0$ and $6A-2B-3C=1$.

Or alternatively: $$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$ If we substitute $s=1$ we get that $$1=A(1+3)(1+2)$$ With $s=-2$ we get that $$1=C(-2-1)(-2+3)$$ And with $s=-3$ we get that $$1=B(-3-1)(-3+2)$$

And here is a video about the partial fraction decomposition, you might find it helpful.

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  • $\begingroup$ how do you know what coefficients to put into A , B and C for the “A+B+C” , “5A+B+2C” , “6A-2B-3C” ? $\endgroup$
    – user185692
    Jul 22, 2018 at 22:02
  • $\begingroup$ @user185692 What do you mean? Would you like to ask the way I've got the 3 equation or the way you can solve it? $\endgroup$
    – Botond
    Jul 22, 2018 at 22:04
  • $\begingroup$ The way you have gotten the 3 equations. $\endgroup$
    – user185692
    Jul 22, 2018 at 22:07
  • $\begingroup$ @user185692 So, after expanding everything, we get that $1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$. Now we want to make the right-hand side nicer, to look like a "normal" polynomial, something like the $ax^2+bx+c$. And if I didn't mess it up, we get that $1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$. The $2$ sides look really different, right? But they should be equal for every $s$! So how can it be? If the right-hand side looks exactly like the left-hand side. So, how many $s^2$ do we have on the right? $A+B+C$. And on the left? $0$. So they should be equal: $A+B+C=0$, and so on. Is it clear now? $\endgroup$
    – Botond
    Jul 22, 2018 at 22:11
  • $\begingroup$ Yes!! Thanks a lot .. $\endgroup$
    – user185692
    Jul 22, 2018 at 22:32
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$$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$ Substitute $k=s+2$ $$F(k) = \frac{1}{(k-3)(k+1)k}$$

$$F(k) = \frac{1}{(k-3)}\left(\frac 1 {(k+1)k} \right)$$ $$F(k) = \frac{1}{(k-3)}\left(\frac 1 k-\frac 1 {k+1} \right)$$ $$F(k) = \frac{1}{(k-3)}\frac 1 k-\frac{1}{(k-3)}\frac 1 {(k+1)} $$ $$F(k) = \frac 13 \left (\frac{1}{(k-3)}-\frac 1 k\right )-\frac 14\left(\frac{1}{(k-3)}-\frac 1 {(k+1)} \right )$$ $$F(k) = \frac 1 {12} \frac{1}{(k-3)}-\frac 1{3k}+\frac 14\frac 1 {(k+1)} $$ Substitute back $k=s+2$ $$X(s)=-\frac 1{3(s+2)}+\frac 1{4(s+3)}+\frac 1 {12(s-1)}$$ $$x(t)=-\frac 1{3}e^{-2t}+\frac 1{4}e^{-3t}+\frac 1 {12}e^{t}$$ It's a bit longer but i find this method easier

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