Question

differential equation using classical method. assume zero condition dx/dt+7x=5cos2t
anybodies can help me?

Answer 1

where are you stuck?

Answer 2

i don't have no idea about that..so i asking the whole solution

Answer 3

i mean i don't have idea

Answer 4

and this is the link for integral by part on the paper http://www.wolframalpha.com/input/?i=integral%285cos%282t%29*e^%287t%29+dt%29

Answer 5

thank you..3x

Answer 6

ok

Answer 7

if you have time? pls. answer my course work no. 3 B to E

Answer 8

It's not the way to help. I can solve it, it's not too hard for me, but what do you get after then? you need study, right? try once, i will correct you if you have mistake.

Answer 9

it's right,i will study to know all about,promse,

Answer 10

@jayven, "classical method?" Is that referring to the Laplace transform?

Answer 11

yup..

Answer 12

Well @Loser66 walked you through part (a) by finding the integrating factor, not the Laplace transform.

Answer 13

need your help @SithsAndGiggles XD

Answer 14

@SithsAndGiggles ,,,
can you give a solutions using laplace transform or classical methods?

Answer 15

and assume zero condition?

Answer 16

\[\frac{dx}{dt}+7x=5\cos2t\]
Take the Laplace of both sides:
\[\mathcal{L}\left\{\frac{dx}{dt}+7x\right\}=\mathcal{L}\left\{5\cos2t\right\}\]
The transform is a linear operator, so
\[\mathcal{L}\left\{\frac{dx}{dt}\right\}+7\mathcal{L}\left\{x\right\}=5\mathcal{L}\left\{\cos2t\right\}\]
The Laplace of a derivative is as follows: \(\mathcal{L}\{y'\}=s\mathcal{L}\{y\}-y(0)\). You assume zero initial conditions, so \(y(0)=0\).
\[s\mathcal{L}\left\{x\right\}-0+7\mathcal{L}\left\{x\right\}=5\mathcal{L}\left\{\cos2t\right\}\]
The Laplace transform \(\cos at\) is \(\dfrac{s}{s^2+a^2}\). (Refer to a transform table, or evaluate the integral from the definition of the transform. The latter would be good practice.)
\[\begin{align*}s\mathcal{L}\left\{x\right\}+7\mathcal{L}\left\{x\right\}&=5\frac{s}{s^2+2^2}\\
(s+7)\mathcal{L}\left\{x\right\}&=5\frac{s}{s^2+2^2}\\
\mathcal{L}\left\{x\right\}&=\frac{5s}{(s+7)\left(s^2+4\right)}
\end{align*}\]
Partial fraction decomposition for the right side:
\[\begin{align*}\frac{5s}{(s+7)\left(s^2+4\right)}&=\frac{A}{s+7}+\frac{Bs+C}{s^2+4}\\
5s&=A\left(s^2+4\right)+(Bs+C)(s+7)\\
5s&=(A+B)s^2+(7B+C)s+(4A+7C)
\end{align*}\]
You get the system,
\[\begin{cases}A+B=0\\\\7B+C=5\\\\4A+7C=0\end{cases}\]
You'll find that \(A=-\dfrac{35}{53},B=\dfrac{35}{53}, C=\dfrac{20}{53}\).
\[\begin{align*}\mathcal{L}\left\{x\right\}&=-\frac{35}{53(s+7)}+\frac{35s+20}{53\left(s^2+4\right)}
\end{align*}\]
Now take the inverse transform:
\[\begin{align*}\mathcal{L}^{-1}\left\{\mathcal{L}\left\{x\right\}\right\}&=\mathcal{L}^{-1}\left\{-\frac{35}{53(s+7)}+\frac{35s+20}{53\left(s^2+4\right)}\right\}
\end{align*}\]
The inverse transform is also a linear operator, so
\[\begin{align*}x&=-\frac{35}{53}\mathcal{L}^{-1}\left\{\frac{1}{s+7}\right\}+\frac{35}{53}\mathcal{L}^{-1}\left\{\frac{s}{s^2+4}\right\}+\frac{20}{53}\mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\}
\end{align*}\]
Refer again to a transform table:
\[\begin{align*}x&=-\frac{35}{53}e^{-7t}+\frac{35}{53}\cos 2t+\frac{20}{53}\sin2t
\end{align*}\]

Answer 17

this is the final answer for laplace?

Answer 18

A/MA/ZING

Answer 19

@SithsAndGiggles
how about my course work no.3 B to E can you also give a solution?

Answer 20

How about no? This is your homework. You already have an idea of how to approach it. You're welcome.