Finding the Solution to x' = 5e^t + y, y' = -4x

  • Thread starter tandoorichicken
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In summary, the system of differential equations given can be solved using the general solution x(t) = c1e^(4t) + c2e^(-t) + 5/20, y(t) = c1e^(4t) - c2e^(-t) - 5/4, where c1 and c2 are arbitrary constants. To find the particular solution, initial conditions must be provided to solve for the constants. This system can be solved numerically and represents the position and velocity of a particle at any given time. It has various real-world applications in physics and engineering.
  • #1
tandoorichicken
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The problem goes like this:

"Show that [itex]x = \sin(2t) + e^t, y = 2\cos(2t) - 4e^t[/itex] is a solution of
[tex]x' = 5e^t + y, y' = -4x [/tex]
and find a family of solutions of this system.

I already did the 'showing' part. I am having trouble finding the general solution for this system. Can someone help me?
 
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  • #2
Have you looked at the second derivative of x?
 
  • #3


To find the general solution for this system, we can use the method of undetermined coefficients. We start by assuming that the solutions have the form x = Ae^t + B\sin(2t) + C\cos(2t) and y = De^t + E\sin(2t) + F\cos(2t), where A, B, C, D, E, and F are constants to be determined.

Next, we substitute these expressions into the given system of equations and equate the coefficients of each term. This will give us a system of equations that we can solve for the unknown coefficients.

For the first equation x' = 5e^t + y, we have:
Ae^t + 2B\cos(2t) - 2C\sin(2t) = 5e^t + De^t + E\cos(2t) - F\sin(2t)

Equating the coefficients of e^t, we get A + D = 5.
Equating the coefficients of \cos(2t), we get 2B + E = 0.
Equating the coefficients of \sin(2t), we get -2C - F = 1.

Similarly, for the second equation y' = -4x, we have:
De^t + 2E\cos(2t) - 2F\sin(2t) = -4(Ae^t + B\sin(2t) + C\cos(2t))

Equating the coefficients of e^t, we get D - 4A = 0.
Equating the coefficients of \cos(2t), we get 2E + 4C = 0.
Equating the coefficients of \sin(2t), we get -2F - 4B = -4.

Solving this system of equations, we get A = -\frac{1}{5}, B = \frac{3}{10}, C = \frac{1}{10}, D = -\frac{4}{5}, E = -\frac{3}{5}, and F = \frac{1}{5}.

Therefore, the general solution for this system is:
x = -\frac{1}{5}e^t + \frac{3}{10}\sin(2t) +
 

1. What is the general solution to the system of differential equations x' = 5e^t + y, y' = -4x?

The general solution to this system of differential equations is given by x(t) = c1e^(4t) + c2e^(-t) + 5/20, y(t) = c1e^(4t) - c2e^(-t) - 5/4, where c1 and c2 are arbitrary constants.

2. How can we determine the particular solution to this system of differential equations?

To determine the particular solution, we need initial conditions for both equations. These initial conditions can be used to solve for the values of the arbitrary constants c1 and c2, which will then give us the specific solution for the system.

3. Can this system of differential equations be solved numerically?

Yes, this system of differential equations can be solved numerically using methods such as Euler's method or the Runge-Kutta method.

4. How can we interpret the behavior of the solution to this system of differential equations?

The solution to this system of differential equations represents the position and velocity of a particle at any given time, with x(t) representing the position and y(t) representing the velocity. The constants c1 and c2 determine the specific behavior of the particle, such as its initial position and velocity.

5. Are there any real-world applications for this system of differential equations?

Yes, this system of differential equations can be used to model a variety of physical systems, such as the motion of a spring-mass system or an electrical circuit. It can also be used in physics and engineering to study the behavior of particles and systems in different scenarios.

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