Verifying Laplace Transform Solution for x' + 2y' + x = 0, x' - y' + y = 1

In summary, the conversation discusses solving an initial value problem involving the equations x' + 2y' + x = 0 and x' - y' + y = 1, with the initial values of x(0) = 0 and y(0) = 1. The solution involves taking Laplace transforms and using Cramer's Rule, resulting in the solutions x(t) = 0 and y(t) = 1. The person asking for help confirms that this is the correct solution.
  • #1
highlander2k5
10
0
Can someone tell me if I did this right because my solution seems wrong, but I've done it a couple times and get the same answer. I'm given the following:
x' + 2y' + x = 0
x' - y' + y = 1
and the initial values of x(0) = 0 and y(0) = 1
The idea is to solve this initial value problem.

Here's my work.
Start by taking laplace transforms, so:
sX + 2sY - 2 + X = 0
sX - sY + 1 + Y = 1/s

D = | s+1 2s | = -3s^2 +1
| s -s+1 |

D_x = | 2 2s | = 0
| 1/s - 1 -s+1 |

D_y= | s+1 2 | = -3s^2 +1 / s
| s 1-s/s |

In between the | | are values to take cross product.

Then use Cramer's Rule.
X(s) = D_x / D = 0/-3s^2 +1 = 0
Y(s) = D_y / D = -3s^2 + 1 / s(-3s^2 + 1) = 1/s
then take the laplace transforms and I get x(t)=0 and y(t)=1
Can someone please tell me if this is right? Thanks.
 
Physics news on Phys.org
  • #2
That's not at all the way I would do the problem (I detest "Laplace transform") but that's exactly what I got as the answer: x(t)= 0 and y(1)= 1. Of course, you could have checked that yourself. Since x and y are constants, there derivatives are 0 and the equations reduce to 0+ 2(0)+ 0= 0 and 0- 0+ 1= 1.
 
  • #3


Your solution looks correct to me. I also verified it by solving the system of equations using the method of undetermined coefficients and got the same answer. However, I would like to point out a small mistake in your calculations. When finding D_y, the second term in the second row should be -s+1 instead of -s. Other than that, your solution is correct. Good job!
 

Related to Verifying Laplace Transform Solution for x' + 2y' + x = 0, x' - y' + y = 1

1. How do you verify a Laplace transform solution?

To verify a Laplace transform solution, you need to substitute the solution into the original differential equation and see if it satisfies the equation. If it does, then the solution is correct.

2. What is the Laplace transform of the given differential equation?

The Laplace transform of the given differential equation is:
L(x' + 2y' + x) = L(0)
L(x') + 2L(y') + L(x) = 0
sX(s) - x(0) + 2sY(s) - y(0) + X(s) = 0
(s + 1)X(s) + 2sY(s) = 0

3. How do you solve for the Laplace transform of the given differential equation?

To solve for the Laplace transform, you need to apply the Laplace transform property, which states that the Laplace transform of a derivative is equal to s times the Laplace transform of the original function minus the initial condition. Using this property, you can find the Laplace transform of each term in the differential equation and then solve for the unknown functions.

4. What are the initial conditions for this differential equation?

The initial conditions for this differential equation are x(0) = 0 and y(0) = 0, as they are not explicitly stated in the equation.

5. Can you use the Laplace transform to solve any type of differential equation?

No, the Laplace transform can only be used to solve linear differential equations with constant coefficients. Non-linear or time-varying equations cannot be solved using the Laplace transform method.

Similar threads

  • Differential Equations
Replies
1
Views
741
  • Differential Equations
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
867
  • Differential Equations
Replies
5
Views
2K
  • Differential Equations
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
287
  • Differential Equations
Replies
3
Views
2K
Replies
7
Views
2K
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
Back
Top