System of Differential Equations

In summary, the conversation discusses a problem in which a system of two differential equations must be converted to a second order equation in y. The attempt at a solution involves finding the roots of the equation and using initial values to determine the values of x and y. The correct solution involves finding the values of y and then using the relation y'(t)=-4x(t) to find the values of x. Ultimately, the solution for y(t) is (5/2)e^(-4t)+(-3/2)e^(4t) and the solution for x(t) is (5/2)e^(-4t)-(-3/2)e^(4t).
  • #1
Shelnutt2
57
0

Homework Statement



So my friend asked me for help because he assumed having a math degree meant I knew math :rolleyes:

[PLAIN]http://courses.webwork.maa.org:8080/wwtmp/equations/0e/c5a05957810916cfdff379ee0642fc1.png

(dx/dt=−4y and dy/dt = −4x)

Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x 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 and y(0) = 1, what are x and y?



Homework Equations



y(t)=C1e^(at)cos(Bt) + C2e^(at)sin(Bt)

The Attempt at a Solution



So I've tried a few approaches but I've failed. This is the approach I got the farthest with.


d^2y/d^2t = -4

you can then say r^2 + 4 = 0

b^2-4ac = 0 - 4*4 = -16

r = 0 +/- sqrt(-16) / 2 = +/- 2i
y = Ae^(0t)cos(2t) + Be^(0t)sin(2t)
y=Acos(2t)+Bsin(2t)
y' =2Bcos(2t) + 2Asin(2t)

If I solve for x(t) I end up with the same equation as y(t).
If you set up for x(0)=4 and y(0)=1, you end up with:

1=A
4=A?
I know this is easy but I'm just not seeing it. Any help would be appreciated. Thanks
 
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  • #2
Shelnutt2 said:
This is the approach I got the farthest with.


d^2y/d^2t = -4

No, [itex]\frac{d}{dx}\left(\frac{dy}{dt}\right)=-4[/itex] but [tex]\frac{d^2y}{dt^2}=\frac{d}{dt}\left(-4x\right)=-4\frac{dx}{dt}[/itex]
 
  • #3
gabbagabbahey said:
No, [itex]\frac{d}{dx}\left(\frac{dy}{dt}\right)=-4[/itex] but [tex]\frac{d^2y}{dt^2}=\frac{d}{dt}\left(-4x\right)=-4\frac{dx}{dt}[/itex]


Ah ok, I see.

Now I get [itex]\frac{-1}{4}\frac{d^2y}{dt^2}[/itex] = -4y
[itex]\frac{-1}{4}\frac{d^2y}{dt^2}[/itex] +4y = 0
so,
[itex]\frac{-1}{4}r^2 + 4y = 0 [/itex]

y(t) = C1 e^(4t) + C2e^(-4t)

Now doing the same for x(t), I end up with the same equation of

x(t) = C1 e^(4t) + C2e^(-4t),

however given my initial values that can't be.
 
  • #4
Shelnutt2 said:
Ah ok, I see.

Now I get [itex]\frac{-1}{4}\frac{d^2y}{dt^2}[/itex] = -4y
[itex]\frac{-1}{4}\frac{d^2y}{dt^2}[/itex] +4y = 0
so,
[itex]\frac{-1}{4}r^2 + 4y = 0 [/itex]

y(t) = C1 e^(4t) + C2e^(-4t)

Good..

Now doing the same for x(t), I end up with the same equation of

x(t) = C1 e^(4t) + C2e^(-4t),

however given my initial values that can't be.

[itex]x(t)[/itex] is related to [itex]y(t)[/itex] via [itex]y'(t)=-4x(t)[/itex], once you've found [itex]y(t)[/itex], you need only differentiate it once and divide by -4 to obtain [itex]x(t)[/itex]
 
  • #5
gabbagabbahey said:
Good..



[itex]x(t)[/itex] is related to [itex]y(t)[/itex] via [itex]y'(t)=-4x(t)[/itex], once you've found [itex]y(t)[/itex], you need only differentiate it once and divide by -4 to obtain [itex]x(t)[/itex]

Thank you for leading me down the right path! I got it now :)

The solution was
y(t) = (5/2)e^(-4t)+(-3/2)e^(4t)
x(t) =(5/2)e^(-4t)-(-3/2)e^(4t)
 

Related to System of Differential Equations

1. What is a system of differential equations?

A system of differential equations is a set of equations that involves the rates of change of multiple variables. These equations are often used to model complex systems and describe how the variables are related to each other over time.

2. How is a system of differential equations solved?

A system of differential equations can be solved analytically or numerically. Analytical solutions involve finding a general solution that satisfies all of the equations, while numerical solutions use algorithms to approximate the solutions.

3. What are some real-life applications of systems of differential equations?

Systems of differential equations are used in a wide range of fields, including physics, engineering, economics, and biology. They can be used to model population growth, chemical reactions, and fluid dynamics, among other things.

4. What is the difference between ordinary and partial differential equations?

Ordinary differential equations involve only one independent variable, while partial differential equations involve multiple independent variables. Systems of differential equations can include both ordinary and partial equations.

5. Are there any limitations to using systems of differential equations?

One limitation of systems of differential equations is that they can be difficult to solve analytically, especially for complex systems. Additionally, the accuracy of numerical solutions can be affected by the choice of algorithm and initial conditions.

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