Use laplace transform to solve the given system

$a)\frac{dx}{dt}-2x-\frac{dx}{dt}-y=6{e}^{3t}$

$2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6{e}^{3t}x(0)=3y(0)=0$

amanf
2021-01-06
Answered

Use laplace transform to solve the given system

$a)\frac{dx}{dt}-2x-\frac{dx}{dt}-y=6{e}^{3t}$

$2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6{e}^{3t}x(0)=3y(0)=0$

You can still ask an expert for help

asked 2021-01-27

Prove $F\left(s\right)=\frac{1}{s-2}\text{}\text{then}\text{}f\left(t\right)$ is

a)${e}^{2t}u\left(t\right)$

b)$u(t+2)$

c)$u(t-2)$

d)${e}^{-2t}u\left(t\right)$

a)

b)

c)

d)

asked 2022-09-24

I am trying to do a Laplace transform of the following equation:

T1⋅V′(t)+V(t)=K⋅P(t)+Va

The purpose is to solve V(s) equation knowing that V(0)=Va. I followed all the transformation rules and I got the following result:

V(s)=K⋅P(s)s⋅(1+T1)+Vas

But, according to the correction, the expected result is :

V(s)=K⋅P(s)(1+s⋅T1)+Vas

What have I done wrong?

T1⋅V′(t)+V(t)=K⋅P(t)+Va

The purpose is to solve V(s) equation knowing that V(0)=Va. I followed all the transformation rules and I got the following result:

V(s)=K⋅P(s)s⋅(1+T1)+Vas

But, according to the correction, the expected result is :

V(s)=K⋅P(s)(1+s⋅T1)+Vas

What have I done wrong?

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Applications of Differential and Difference Equations

Solve$({D}^{2}-2D+1)y=x{e}^{x}\mathrm{sin}x$

Solve

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Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.)

$L\{{e}^{-t}\cdot {e}^{t}\mathrm{cos}\left(t\right)\}$

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Find the general solution of the given differential equation. $y\u201d+{y}^{\prime}-6y=12{e}^{3t}+12{e}^{-2t}$

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A steel ball of mass 4-kg is dropped from rest from the top of a building. If the air resistance is 0.012v and the ball hits the ground after 2.1 seconds, how tall is the building? Answer in four decimal places.