Problems with Laplace Transforms

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Homework Help Overview

The problem involves a system of differential equations describing the motion of a particle along a plane curve, specifically given by the equations \(\frac{dy}{dt} + 2x=\sin 2t\) and \(\frac{dx}{dt} - 2y=\cos 2t\). The initial conditions are \(x(0)=1\) and \(y(0)=0\), and the goal is to show that the particle moves along the curve defined by \(4x^2+4xy+5y^2=4\) using Laplace transforms.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss applying the Laplace transform to the given differential equations and express uncertainty about the correctness of their transformations and subsequent algebraic manipulations. There are questions regarding the formatting of equations and the interpretation of terms in the context of Laplace transforms.

Discussion Status

Some participants have identified mistakes in their calculations and are revisiting their simultaneous solutions for the transformed variables \(X\) and \(Y\). There is ongoing clarification regarding the application of the Laplace transform and the handling of initial conditions.

Contextual Notes

Participants note issues with formatting equations correctly and the potential confusion arising from mixing different notation styles. There is also a mention of needing to apply the inverse Laplace transform, with some uncertainty about how to handle constants in that context.

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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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