Question

The coordinates (x,y) of a particle moving along a plane curve at any time t are given by y'+ 2x = sin 2t , x' -2 y = cost (t > 0) If at t=0, x=1 and y=0 Find the equation of the curve the particle is moving in terms of x and y by Laplace transformation?

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

WORKING PROCEDURE

The procedure stated below is applied to solve Simultaneous linear differential equations with constant coefficients by Laplace transform method :

STEP 1 : First the Laplace Transform of both sides of the given linear differential equation

STEP 2 : Use the given initial conditions

STEP 3 : Express Laplace transformation operator of y in terms of s

STEP 4 : Express this function of s into partial fractions

STEP 5 : Take the inverse transform in both sides

STEP 6 : Then obtain value of x and y as a function of t satisfying the given initial conditions.

STEP 7 : Eliminating T from the obtained value of x & y is the equation desired

FORMULA TO BE IMPLEMENTED

LAPLACE TRANSFORMS

The Laplace Transforms of f(t) , denoted by L{f(t)} and defined as :

$\sf{\bar{f}(s) = L\{f(t)\} =\displaystyle \int\limits_{0}^{\infty} e^{-st} f(t)\, dt}$

CALCULATION

It is given that

${y}^{} {'} + 2x = \sin 2t \: \: \: ....(1)$

${x}^{} {'} - 2y = \cos 2t \: \: \: ....(2)$

With initial conditions

$t= 0 \: , \: x= 1 \: , \: y= 0$

Taking Laplace transforms of Equation (1) we get

$\displaystyle \: \{s \bar{y} - y(0) \} + 2 \bar{x} = \frac{2}{ {s}^{2} + {2}^{2} } \: \:$

$\implies \: \displaystyle \: 2 \bar{x} + s \bar{y}= \frac{2}{ {s}^{2} + 4 } \: \: \: \: ....(3)$

Again taking Laplace Transforms of Equation (2) we get

$\displaystyle \: \{s \bar{x} - x(0) \} - 2 \bar{y} = \frac{s}{ {s}^{2} + {2}^{2} } \: \:$

$\implies \: \displaystyle \: s \bar{x} - 2 \bar{y} = \frac{s}{ {s}^{2} + {2}^{2} } + 1 \: \: \: \: .....(4)$

Multiplying Equation (3) by s and Equation (4) by 2 and subtracting we get

$\displaystyle \: ( {s}^{2} + 4) \bar{y} = - 2$

$\implies \: \displaystyle \:\bar{y} = - \frac{2}{ ( {s}^{2} + 4) }$

On Inversion

$\displaystyle \: y = - 2 {L}^{ - 1} \bigg( \: \frac{1}{ {s}^{2} + 4} \bigg) = - \sin 2t$

From Equation (3) we get

$2x = \sin 2t - {y}^{} {'}$

$\implies \: 2x = \sin2t + 2 \cos2t$

Squaring both sides we get

$\implies \: 4 {x}^{2} = {(\sin2t + 2 \cos2t)}^{2} ...(5)$

Also

$4xy =( \sin2t + 2 \cos2t) \times ( - 2\sin2t)$

$\implies \: 4xy = - 2( { \sin}^{2}2t + 2 \cos2t\sin2t) \: \: ...(6)$

Also

$5 {y}^{2} = 5 \: {\sin}^{2} 2t ...(7)$

Equation (5) + (6) + (7) gives

$4 {x}^{2} + 4xy + 5 {y}^{2}$

$= { \sin}^{2} 2t + 4 \sin2t \: \cos2t + 4 { \cos}^{2} 2t - 2 { \sin}^{2}2t - 4 \cos2t\sin2t + 5 \: {\sin}^{2} 2t$

$= 4{ \sin}^{2} 2t +4 { \cos}^{2} 2t$

$= 4 \: ({ \sin}^{2} 2t + { \cos}^{2} 2t )$

$= 4 \times 1$

$= 4$

So

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

RESULT

The equation of the curve the particle moving in terms of x and y by Laplace transformation is

$\boxed{ \: 4 {x}^{2} + 4xy + 5 {y}^{2} = 4 \: \: }$