Question

solve the following equation using la place trans form:
3*dx/dt -4x =sin2t , given t=0 and x=1/3

Answer 1

heey i correct it just look at it again

Answer 2

\[\begin{align*}3x'-4x&=\sin2t\\
\mathcal{L}\left\{3x'-4x\right\}&=\mathcal{L}\left\{\sin2t\right\}\\
3\mathcal{L}\left\{x'\right\}-4\mathcal{L}\left\{x\right\}&=\mathcal{L}\left\{\sin2t\right\}\\
3\left(s\mathcal{L}\left\{x\right\}-x(0)\right)-4\mathcal{L}\left\{x\right\}&=\frac{2}{s^2+4}\\
(3s-4)\mathcal{L}\left\{x\right\}-1&=\frac{2}{s^2+4}\\
\mathcal{L}\left\{x\right\}&=\frac{s^2+6}{(s^2+4)(3s-4)}
\end{align*}\]
Does everything make sense so far?

Answer 3

y u dont apply the conditions ??

Answer 4

I did. \(x(0)=\dfrac{1}{3}\). In the fourth line of the equation, I immediately distribute the 3, which leads to the fifth line.
\[\begin{align*}
3\left(s\mathcal{L}\left\{x\right\}-x(0)\right)-4\mathcal{L}\left\{x\right\}&=\frac{2}{s^2+4}\\
3s\mathcal{L}\left\{x\right\}-3x(0)-4\mathcal{L}\left\{x\right\}&=\frac{2}{s^2+4}\\

(3s-4)\mathcal{L}\left\{x\right\}-3\left(\frac{1}{3}\right)&=\frac{2}{s^2+4}\\
(3s-4)\mathcal{L}\left\{x\right\}-1&=\frac{2}{s^2+4}\\
(3s-4)\mathcal{L}\left\{x\right\}&=\frac{2}{s^2+4}+1\\
(3s-4)\mathcal{L}\left\{x\right\}&=\frac{2}{s^2+4}+\frac{s^2+4}{s^2+4}\\
(3s-4)\mathcal{L}\left\{x\right\}&=\frac{s^2+6}{s^2+4}\\
\mathcal{L}\left\{x\right\}&=\frac{s^2+6}{(s^2+4)(3s-4)}
\end{align*}\]

Answer 5

Do you think you can take care of the inverse transform?

Answer 6

im not sure i tried many times but i could not

Answer 7

can u help me

Answer 8

The first step would be to decompose into partial fractions:
\[\begin{align*}
\frac{s^2+6}{(s^2+4)(3s-4)}&=\frac{As+B}{s^2+4}+\frac{C}{3s-4}\\
s^2+6&=(As+B)(3s-4)+C(s^2+4)\\
s^2+6&=3As^2-4As+3Bs-4B+Cs^2+4C\\
s^2+6&=(3A+C)s^2+(3B-4A)s-4(B-C)
\end{align*}\]
So you have
\[\begin{cases}3A+C=1\\3B-4A=0\\-4(B-C)=6\end{cases}~~\Rightarrow~~\begin{cases}A=-\dfrac{3}{26}\\\\
B=-\dfrac{2}{13}=-\dfrac{4}{26}\\\\C=\dfrac{35}{26}\end{cases}\]
So,
\[\mathcal{L}\{x\}=-\frac{3s+4}{26(s^2+4)}+\frac{35}{26(3s-4)}\]
Take the inverse transform of both sides:
\[\begin{align*}
x&=\mathcal{L}^{-1}\left\{-\frac{3s+4}{26(s^2+4)}+\frac{35}{26(3s-4)}\right\}\\
x&=\mathcal{L}^{-1}\left\{-\frac{3s+4}{26(s^2+4)}\right\}+\mathcal{L}^{-1}\left\{\frac{35}{26(3s-4)}\right\}\\
x&=-\frac{1}{26}\mathcal{L}^{-1}\left\{\frac{3s+4}{s^2+4}\right\}+\frac{35}{26}\mathcal{L}^{-1}\left\{\frac{1}{3s-4}\right\}\\
x&=-\frac{1}{26}\mathcal{L}^{-1}\left\{\frac{3s}{s^2+4}\right\}-\frac{1}{26}\mathcal{L}^{-1}\left\{\frac{4}{s^2+4}\right\}+\frac{35}{26}\mathcal{L}^{-1}\left\{\frac{1}{3s-4}\right\}\\
x&=-\frac{3}{26}\color{red}{\mathcal{L}^{-1}\left\{\frac{s}{s^2+4}\right\}}-\frac{1}{13}\color{red}{\mathcal{L}^{-1}\left\{\frac{2}{s^2+4}\right\}}+\frac{35}{26}\mathcal{L}^{-1}\left\{\frac{1}{3s-4}\right\}\end{align*}\]
The two red inverse transforms may be done immediately. The third involves some intermediate steps.
\[\begin{align*}\mathcal{L}^{-1}\left\{\frac{1}{3s-4}\right\}&=\frac{1}{3}\mathcal{L}^{-1}\left\{\frac{1}{s-\dfrac{4}{3}}\right\}
\end{align*}\]
You have a transform of the form \(F(s-c)\), which means its inverse transform will have the form \(\large e^{ct}f(t)\).
\[\frac{1}{3}\mathcal{L}^{-1}\left\{\frac{1}{s-\dfrac{4}{3}}\right\}=\frac{1}{3}e^{4/3~t}\mathcal{L}^{-1}\left\{\frac{1}{s}\right\}=\frac{e^{4/3 ~t}}{3}\]
Put it all together.