Subject to

y(0)=0

y'(0)=0

Caelan
2021-09-03
Answered

Solve ${y}^{\prime}+y=4\delta (t-2\pi )$

Subject to

y(0)=0

y'(0)=0

Subject to

y(0)=0

y'(0)=0

You can still ask an expert for help

timbalemX

Answered 2021-09-04
Author has **108** answers

Step 1

According to the given information, it is required to solve the given differential equation.

${y}^{\prime}+y=4\delta (t-2\pi )$

Subject to y(0)=0 , y'(0)=0

Step 2

Use laplace transform to solve the above equation.

${y}^{\prime}+y=\delta (t-2\pi )$

$L\left({y}^{\prime}\right)+L\left(y\right)=4L\left(\delta (t-2\pi )\right)$

$sY\left(s\right)-y\left(0\right)+Y\left(s\right)=4\left[{e}^{-2\pi s}\right](L\left(\delta (t-c)\right)={e}^{-cs})$

$(s+1)Y\left(s\right)=4\left[{e}^{-2\pi s}\right]$

$Y\left(s\right)=\frac{4\left[{e}^{-2\pi s}\right]}{s+1}$

$y\left(t\right)=4{L}^{-1}\left(\frac{\left[{e}^{-2\pi s}\right]}{s+1}\right)$

$y\left(t\right)=4\left[{u}_{2\pi}\left(t\right){e}^{-(t-2\pi )}\right](L\left({u}_{c}\left(t\right)f(t-c)\right)={e}^{-cs}F\left(s\right))$

According to the given information, it is required to solve the given differential equation.

Subject to y(0)=0 , y'(0)=0

Step 2

Use laplace transform to solve the above equation.

asked 2022-04-27

y"+4y=${\int}_{1}^{0}0\le t\le t\le \infty $

y(0)=1

y'(0)=0

asked 2020-12-24

Find the laplace transform of the following function

$f(t)=t{u}_{2}(t)$

asked 2022-01-18

Solve: $y{}^{\u2033}+2{y}^{\prime}-3y=0$

asked 2021-02-16

Solution of I.V.P for harmonic oscillator with driving force is given by Inverse Laplace transform

1)

2)

3)

4)

asked 2021-09-19

Do the laplace transform of

asked 2022-05-15

Im clueless on how to solve the following question...

$x{e}^{y}\frac{dy}{dx}={e}^{y}+1$

What i've done is...

$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{x{e}^{e}};\frac{dy}{dx}-\frac{1}{x{e}^{e}}=\frac{1}{x}$

Find the integrating factor..

$v(x)={e}^{P(x)};whereP(x)=\int p(x)dx\Rightarrow P(x)=\int \frac{1}{x}dx=ln|x|\phantom{\rule{0ex}{0ex}}v(x)={e}^{P(x)}={e}^{ln|x|}=x;\phantom{\rule{0ex}{0ex}}y=\frac{1}{v(x)}\int v(x)q(x)dx=\frac{1}{x}\int x\frac{1}{x}dx=1+c$

I know I made a mistake somewhere. Would someone advice me on this?

$x{e}^{y}\frac{dy}{dx}={e}^{y}+1$

What i've done is...

$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{x{e}^{e}};\frac{dy}{dx}-\frac{1}{x{e}^{e}}=\frac{1}{x}$

Find the integrating factor..

$v(x)={e}^{P(x)};whereP(x)=\int p(x)dx\Rightarrow P(x)=\int \frac{1}{x}dx=ln|x|\phantom{\rule{0ex}{0ex}}v(x)={e}^{P(x)}={e}^{ln|x|}=x;\phantom{\rule{0ex}{0ex}}y=\frac{1}{v(x)}\int v(x)q(x)dx=\frac{1}{x}\int x\frac{1}{x}dx=1+c$

I know I made a mistake somewhere. Would someone advice me on this?

asked 2022-01-19

Solve $\frac{{d}^{2}\theta}{{dt}^{2}}+g\mathrm{sin}\theta =0$