Problems with Laplace Transforms

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The discussion focuses on solving a set of differential equations using Laplace transforms to demonstrate the motion of a particle along a specified curve. The equations provided are transformed, leading to expressions for X and Y in terms of s. Participants highlight issues with formatting and the accuracy of the simultaneous solutions for X and Y. Corrections are made regarding the transformation process and the need for consistent notation. The thread emphasizes the importance of careful algebraic manipulation in obtaining the correct results.
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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
 
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Sorry those fractions didn't come properly
 
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=\frac{2}{s<sup>2</sup>+4}

2X+sY=\frac{2}{s<sup>2</sup>+4} -2

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

Applying LT,

sX-2Y=\frac{s<sup>2</sup>+s+4}{s<sup>2</sup>+4} -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
 
LaTeX fixed.
 
Hmpf, I left that as an exercise to the reader.
 
Your simultaneous solutions for X and Y are incorrect. Try again.

Chet
 
faiz4000 said:

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=\frac{2}{s<sup>2</sup>+4}

2X+sY=\frac{2}{s<sup>2</sup>+4} -2



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

Where did the -3 come from?

Applying LT,

sX-2Y=\frac{s<sup>2</sup>+s+4}{s<sup>2</sup>+4} -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?
 
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)
 
  • #10
thanx for your help
 

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