differential equations

**computers** · July 19th 2012, 04:19 PM

Find Inverse Transform of

(s)/(s^2 + a^2)(s^2+b^2), a^2 not equal to b^2, ab is not equal to 0

**BAdhi** · July 19th 2012, 07:47 PM

$\frac{s}{(s^2+a^2)(s^2+b^2)}=\frac{s}{s^2+b^2}. \frac{1}{s^2+a^2}$

take,

$F(s)=\frac{1}{s^2+a^2}$

and

$G(s)=\frac{s}{s^2+b^2}$

now use convolution property of Laplace transform

**computers** · July 21st 2012, 05:32 AM

can u please explain further.

**BAdhi** · July 22nd 2012, 04:36 AM

the convolution property is,

$L\left[ \int_0^tf(\tau)g(t-\tau )d\tau \right] =F(s).G(s)$

where $L[f(t)]=F(s)$ and $L[g(t)]=G(s)$

now if you take things as above post(#2),

when $F(s)=\frac{1}{s^2+a^2}$ we know that corresponding $f(t)$ is $\frac{1}{a}\sin (at)$ (for $Re(s)>0$ )
and when $G(s)=\frac{s}{s^2+b^2}$ we know that corresponding $g(t)$ is $\cos (bt)$ (for $Re(s)>0$ )

then according to the above property,

$L^{-1}\left[F(s).G(s)\right]=\int_0^t\frac{1}{a}\sin (a\tau ).\cos [b(t-\tau )]d\tau$

Can you continue from here?

Of course there are some other methods to solve this question as well....

**Prove It** · July 22nd 2012, 05:13 AM

Originally Posted by computers

Find Inverse Transform of

(s)/(s^2 + a^2)(s^2+b^2), a^2 not equal to b^2, ab is not equal to 0

$\displaystyle \begin{align*} \frac{s}{\left(s^2 + a^2\right)\left(s^2 + b^2\right)} &= \frac{As + B}{s^2 + a^2} + \frac{Cs + D}{s^2 + b^2} \\ \frac{s}{\left(s^2 + a^2\right)\left(s^2 + b^2\right)} &= \frac{(As + B)\left(s^2 + b^2\right) + (Cs + D)\left(s^2 + a^2\right)}{\left(s^2 + a^2\right)\left(s^2 + b^2\right)} \\ s &= (As + B)\left(s^2 + b^2\right) + (Cs + D)\left(s^2 + a^2\right) \\ s &= As^3 + Ab^2s + Bs^2 + Bb^2 + Cs^3 + Ca^2s + Ds^2 + Da^2 \\ 0s^3 + 0s^2 + 1s + 0 &= (A + C)s^3 + (B + D)s^2 + \left(Ab^2 + Ca^2\right)s + Bb^2 + Da^2 \end{align*}$

Equating like coefficients of $\displaystyle \begin{align*} s^3 \end{align*}$ gives $\displaystyle \begin{align*} A + C = 0 \implies C = -A \end{align*}$ .

Equating like coefficients of $\displaystyle \begin{align*} s \end{align*}$ gives

$\displaystyle \begin{align*} Ab^2 + Ca^2 &= 1 \\ Ab^2 - Aa^2 &= 1 \\ A\left(b^2 - a^2\right) &= 1 \\ A &= \frac{1}{b^2 - a^2} \\ C &= \frac{1}{a^2 - b^2} \end{align*}$

Equating like coefficients of $\displaystyle \begin{align*} s^2 \end{align*}$ gives $\displaystyle \begin{align*} B + D = 0 \implies D = -B \end{align*}$ .

Equating like constants gives

$\displaystyle \begin{align*} Bb^2 + Da^2 &= 0 \\ Bb^2 - Ba^2 &= 0 \\ B\left(b^2 - a^2\right) &= 0 \\ B &= 0 \\ D &= 0 \end{align*}$

Therefore we have

$\displaystyle \begin{align*} \frac{s}{\left(s^2 + a^2\right)\left(s^2 + b^2\right)} &= \frac{s}{\left(b^2 - a^2\right)\left(s^2 + a^2\right)} + \frac{s}{\left(a^2 - b^2\right)\left(s^2 + b^2\right)} \\ &= \frac{1}{a^2 - b^2}\left(\frac{s}{s^2 + b^2} - \frac{s}{s^2 + a^2} \right) \end{align*}$

Thus

$\displaystyle \begin{align*} \mathcal{L}^{-1}\left\{ \frac{s}{\left(s^2 + a^2\right)\left(s^2 + b^2\right)} \right\} &= \frac{1}{a^2 - b^2} \, \mathcal{L}^{-1} \left\{ \frac{s}{s^2 + b^2} - \frac{s}{s^2 + a^2} \right\} \\ &= \frac{1}{a^2 - b^2} \left[ \cos{\left( b\,t \right)} - \cos{\left( a\,t \right)} \right] \end{align*}$

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