Problems with Laplace Transforms

[faiz4000]

Homework Statement

The coordinates $(x,y)$ of a particle moving along a plane curve at any time t, are given by

$$\frac{dy}{dt} + 2x=\sin 2t,$$
$$\frac{dx}{dt} - 2y=\cos 2t.$$

If at $t=0$, $x=1$ and $y=0$, using Lapace transform show that the particle moves along the curve

$$4x^2+4xy+5y^2=4$$

Homework Equations

note: Lowercase letters $x$,$y$ are functions of $t$. Uppercase letters $X$,$Y$ are functions of $s$.

The Attempt at a Solution

Apply Laplace Transform on the given equations

$$\frac{dy}{dt} + 2x=\sin 2t~~~~~~~~(1)$$

Applying LT,

$$sY-y(0)+2X=\frac{2}{s^2+4}$$

$$2X+sY= \frac{2}{s^2+4}~~~~~~~~(2)$$

$$\frac{dx}{dt} - 2y=\cos 2t~~~~~~~~(3)$$

Applying LT,

$$sX-2Y=\frac{s^2+s+4}{s^2+4}~~~~~~~~(4)$$

Now Solving $(2)$ and $(4)$ simultaneously,

$$X=-s^3-s^2-4s-4$$

$$Y=2s^2+8$$

Now I have to apply Inverse Laplace transform to get back $x$ and $y$ but i don't know how to get ILT of a constant... Also not sure everything I've done till now is right so please help

 

[faiz4000]

Sorry those fractions didn't come properly

 

[faiz4000]

Solving Simultaneous Differential Equations using Laplace Transform

Homework Statement

The coordinates (x,y) of a particle moving along a plane curve at any time t, are given by

$\frac{dy}{dt}$ + 2x=sin2t,
$\frac{dx}{dt}$ - 2y=cos2t

If at t=0, x=1 and y=0, using Lapace transform show that the particle moves along the curve

4x²+4xy+5y²=4

Homework Equations

note: Lowercase letters x,y are functions of t. Uppercase letters X,Y are functions of s.

The Attempt at a Solution

Apply Laplace Transform on the given equations

$\frac{dy}{dt}$ + 2x=sin2t -1

Applying LT,

sY-y(0)+2X=[itex]\frac{2}{s²+4}[/itex]

2X+sY=[itex]\frac{2}{s²+4}[/itex] -2

$\frac{dx}{dt}$ - 2y=cos2t -3

Applying LT,

sX-2Y=[itex]\frac{s²+s+4}{s²+4}[/itex] -4

Now Solving 2 and 4 simultaneously,

X=-s³-s²-4s-4

Y=2s²+8

Now I have to apply Inverse Laplace transform to get back x and y but i don't know how to get ILT of a constant... Also not sure everything I've done till now is right so please help

 

[Borek]

That's because you are mixing different ways of formatting - tex with BBC. It never works. Be consistent, format whole formula one way (or another).

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[micromass]

LaTeX fixed.

 

[Borek]

Hmpf, I left that as an exercise to the reader.

 

[Chestermiller]

Your simultaneous solutions for X and Y are incorrect. Try again.

Chet

 

[LCKurtz]

Homework Statement

The coordinates (x,y) of a particle moving along a plane curve at any time t, are given by

$\frac{dy}{dt}$ + 2x=sin2t,
$\frac{dx}{dt}$ - 2y=cos2t

If at t=0, x=1 and y=0, using Lapace transform show that the particle moves along the curve

4x²+4xy+5y²=4

Homework Equations

note: Lowercase letters x,y are functions of t. Uppercase letters X,Y are functions of s.

The Attempt at a Solution

Apply Laplace Transform on the given equations

$\frac{dy}{dt}$ + 2x=sin2t -1

Where did the -1 come from?

Applying LT,

sY-y(0)+2X=[itex]\frac{2}{s²+4}[/itex]

2X+sY=[itex]\frac{2}{s²+4}[/itex] -2



$\frac{dx}{dt}$ - 2y=cos2t -3

Where did the -3 come from?

Applying LT,

sX-2Y=[itex]\frac{s²+s+4}{s²+4}[/itex] -4

Now Solving 2 and 4 simultaneously,

X=-s³-s²-4s-4

Y=2s²+8

If the -1 and -3 are supposed to be in there, you didn't transform them. And given the equations you have, how did you get those values for X and Y?

 

[faiz4000]

those were eqn numbers...eqn 1 and eqn 3... I had left space but it didn't reflect in the post...I found my mistake... it was in solving the simultaneous eqns...you actually get X=(s+1)/(s^2+4)
Y=-s/(s^2+4)

 

[faiz4000]

thanx for your help