Question

Consider the system of differential equations dx/dt=−2y dy/dt=−2x. . Convert this system to a second order differential equation in yy by differentiating the second equation with respect to tt and substituting for xx from the first equation. Solve the equation you obtained for y as a function of t; hence find x as a function of t. If we also require x(0)=4 y(0)=3, what are x and y?x(t)= y(t)=

Answer 1

Answer:

Step-by-step explanation:

we have the following differential equations

$\frac{dx}{dt}=-2y\\\frac{dy}{dt}=-2x$

by differentiating the second equation we have

$\frac{d}{dt}(\frac{dy}{dt})=-2\frac{dx}{dt}\\\frac{d^{2}y}{dt^{2}}=-2\frac{dx}{dt}\\\frac{dx}{dt}=\frac{-1}{2}\frac{d^{2}y}{dt^{2}}$

and we replace dx/dt in the first equation

$\frac{-1}{2}\frac{d^{2}y}{dt^{2}}=-2y\\\frac{d^{2}y}{dt^{2}}-4y=0$

and by using the characteristic polynomial

$m^{2}+4=0\\m=\±2i$

the solution is

$y(t)=Acos(2t)+Bsin(2t)$

and to compute x(t) we have

$\frac{dx}{dt}=-2Acos(2t)-2Bsin(2t)\\\\\int dx = \int[-2Acos(2t)-2Bsin(2t)]dt\\\\x(t)=-Asin(2t)+Bcos(2t)$

and if we use x(0)=4 and y(0)=3, we can calculate the constants A and B

$x(0)=B=4\\y(0)=A=3$

I hope this is useful for you

regards

Answer 2

Answer:

x(t) = (13/8)e^(2t) + (19/8)e^(-2t)

y(t) = -(13/2)e^(2t) + (19/2)e^(-2t)

Step-by-step explanation:

Given the system of differential equations:

dx/dt = -2y.................................(1)

dy/dt = -2x.................................(2)

Differentiating (1)

d²x/dt² = -2dy/dt.......................(3)

Using the value dy/dt = -2x from (2) in (3)

d²x/dt² = -2(-2x) = 4x

d²x/dt² - 4x = 0.............................(4)

The auxiliary equation to (4) is

m² - 4 = 0

(m - 2)(m + 2) = 0

m - 2 = 0

=> m = 2

m + 2 = 0

=> m = -2

m = 2, -2

The complimentary solution is therefore

x = C1e^(2t) + C2e^(-2t)............(5)

But dy/dt = -2x

dy/dt = -2C1e^(2t) - 2C2e^(-2t)

Integrating this, we have

y = -4C1e^(2t) + 4C2e^(-2t).......(6)

Applying the condition x(0) = 4

Put x = 4 when t = 0 in (5)

4 = C1 + C2....................................(7)

Applying the condition y(0) = 3

Put y = 3 when t = 0 in (6)

3 = -4C1 + 4C2 .............................(8)

Solve (7) and (8) simultaneously

From (7)

C2 = 4 - C1

Using this in (8)

3 = -4C1 + 4(4 - C1)

3 = -4C1 + 16 - 4C1

-8C1 = -13

C1 = 13/8 ********

But C2 = 4 - C1

=> C2 = 4 - 13/8

C2 = 19/8 *******

Using the values of C1 and C2 in (5) and (6), we have

x = (13/8)e^(2t) + (19/8)e^(-2t)

y = -(13/2)e^(2t) + (19/2)e^(-2t)