# Use laplace transform to solve the given system a) frac{dx}{dt} -2x- frac{dx}{dt}-y =6e^{3t} 2left(frac{dx}{dt}right)-3x+frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0

Question
Laplace transform
Use laplace transform to solve the given system
$$a) \frac{dx}{dt} -2x- \frac{dx}{dt}-y =6e^{3t}$$
$$2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0$$

2021-01-07
Step 1
Given
$$a) \frac{dx}{dt} -2x- \frac{dx}{dt}-y =6e^{3t}$$
$$2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0$$
We use the Laplace transforms initially,
$$L(x(t))=\bar x(s)$$
$$L(y(t))=\bar y(s)$$
$$L(x'(t))=s\bar x(s)-x(0)$$
$$L(y'(t))=s\bar y(s)-y(0)$$
And in the latter stage, we use the known inverse Laplace transforms to get the required solution.
Step 2
So, we have
$$s\bar x(s)-x(0)-2\bar x(s)-s\bar y(s)+y(0)+\bar y(s)=\frac{6}{(s-3)}$$
$$2s\bar x(s)-2x(0)-3\bar x(s)+s\bar x(s)-y(0)-3\bar y(s)=\frac{6}{(s-3)}$$
Simplifying and applying the initial conditions,
$$(s-2)\bar x(s)+(1-s)\bar y(s)=\frac{6}{(s-3)}+3 \dots (1)$$
$$(2s-3)\bar x(s)+(x-3)\bar y(s)=\frac{6}{(s-3)}+3 \dots (2)$$
Using Cramer's rule,
$$\bar x(s)=\frac{\begin{vmatrix}\frac{6}{(s-3)}+3 & 1-s \\\frac{6}{(s-3)}+6 & s-3 \end{vmatrix}}{\begin{vmatrix}s-2 & 1-s \\2s-3 & s-3 \end{vmatrix}}$$
$$=\frac{3s-3-\frac{6(s-2)(1-s)}{s-3}}{(s-2)(s-3)-(2s-3)(1-s)}$$
$$=\frac{(s-1)(9s-21)}{3s^2-10s+9} \dots (3)$$
$$=3-\frac{6}{3s^2-10s+9}$$
$$=3-\frac{2}{\left(s-\frac{10}{6}\right)^2+\left(\sqrt{\frac{8}{6}}\right)^2}$$
$$x(t)=3\delta(t)-6/\sqrt{8}e^{\frac{10t}{6}}\sin\left(\frac{\sqrt{8t}}{6}\right)$$
Where, $$\delta(t)$$ is the delta function
Step 3
Substituting (3) in (2), we get
$$\frac{(2s-3)(s-1)(9s-21)}{3s^2-10s+9}+(s-3)\bar y(s)=\frac{6}{(s-3)}+6$$
$$(s-3)\bar y(s)=\frac{6(s-2)}{(s-3)}-\frac{(2s-3)(s-1)(9s-21)}{3s^2-10s+9}$$
$$\bar y(s)=\frac{6(s-2)}{(s-3)}-\frac{(2s-3)(s-1)(9s-21)}{(s-3)}(3s^2-10s+9)$$
$$\bar y(s)=\frac{-18s^4+159s^3-489s^2+633s-297}{(s-3)^2(3s^2-10s+9)}$$
$$\bar y(s)=-6+\frac{2}{(s-3)^2}+\frac{5-3s}{3s^2-10s+9}$$
$$y(t)=-6\delta(t)+2e^{3t}t-e^{\frac{10t}{6}}\cos\left(\frac{\sqrt8t}{6}\right)$$
Step 4
Conclusion:
$$x(t)=3\delta(t)-\frac{6}{\sqrt8}e^{\frac{10t}{6}}\sin\left(\frac{\sqrt8t}{6}\right)$$
$$y(t)=-6\delta(t)+2e^{3t}t-e^{\frac{10t}{6}}\cos\left(\frac{\sqrt8t}{6}\right)$$
Where, $$\delta(t)$$ is the delta function, is the required solution to the system of equations.

### Relevant Questions

Use the Laplace transform to solve the given system of differential equations.
$$\frac{(d^2x)}{(dt^2)}+\frac{(d^2y)}{(dt^2)}=\frac{t}{2}$$
$$\frac{(d^2x)}{(dt^2)}-\frac{(d^2y)}{(dt^2)}=4t$$
$$x(0) = 5, x'(0) = 0,$$
$$y(0) = 0, y'(0) = 0$$
Find x(t) and Y(t) using Laplace transform. $$\begin{cases} \frac{dx}{dt}=2x-3y \frac{dy}{dt}=y-2x \end{cases}$$

$$x(0)=8 , y(0)=3$$
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?
solve the initial value problem with using laplace
$$Y'+3Y=6e^{-3t}\cos 6t+24$$
$$Y(0)=0$$
ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
$$L(y) = \left[\frac{w}{(s^2 + a^2)(s^2+w^2)}\right]*b$$
Problem 2 Solve the differential equation
$$\frac{d^2y}{dt^2}+a^2y=b \sin(\omega t)$$ where $$y(0)=0$$
and $$y'(0)=0$$
use the Laplace transform to solve the given initial-value problem.
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$
Use the Laplace transform to solve the given initial-value problem.
$$\displaystyle{y}{''}+{8}{y}'+{41}{y}=\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)},\ \ \ \ {y}{\left({0}\right)}={1},\ \ \ \ {y}'{\left({0}\right)}={0}\ \ {y}{\left({t}\right)}=?$$

A system of linear equations is given below.
$$2x+4y=10$$
$$\displaystyle-{\frac{{{1}}}{{{2}}}}{x}+{3}={y}$$
Find the solution to the system of equations.
A. (0, -3)
B. (-6, 0)
C. There are infinite solutions.
D. There are no solutions.

a) $$x' +4x+3y=0 \text{ with } x(0)=0$$
$$y'+3x+4y=2e^t , y(0)=0$$
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{{0}}}^{{{x}}}}{e}^{{\sqrt{{{t}}}}}{\left.{d}{t}\right.}$$