# Using laplace to solve 2 equationsplease check workthanks

• fufufu
In summary, using Laplace transforms, the solutions to the given system of equations are x(t) = (1/3)e^4t - (1/3)e^t and y(t) = e^t.
fufufu

## Homework Statement

use laplace transforms to solve system of two equations
x' = x -4y + e^4t
y' = -x + y -4e^4t
x(0) = 0
y(0) =1

## Homework Equations

i am uncertain about the correct order to solve these problems.. am I taking inverse L.T. at right time and place in order for the prob to check out? thanks for any help

## The Attempt at a Solution

sX(s) - 0 - X(s) = -4Y(s) + 1/s-4
X(s) = -4Y(s)/s-1 + 1/(s-4)(s-1)

sY(s) -1 -Y(s) - 4Y(s)/s-1) = -1/(s-4)(s-1)
Y(s) = 1/(s-1)(s-3)(s-4) + 1/s-3

1 = A(s-1)(s-3) + B(s-4)(s-3) + C(s-4)(s-1)
after letting s = 1, 3 and 4 i get
A=1/3, B=1/6 and C=-1/2

Y(s) = 1/3(1/s-4) + 1/6(1/s-1) -1/2(1/s-3)

y(t) = (1/3)e^4t + (1/6)e^t - (1/2)e^3t

y(t) = e^t

X(s) = -4Y(s)(1/s-1) + 1/(s-4)(s-1)
after using partial fractions on rightmost term on RHA and inserting y(t) for other term i get

X(s) = -4e^t(1/s-1) + 1/(s-4)(s-1)

x(t) = -4e^t(e^t) + (1/3)(1/s-4) - (1/3)(1/s-1)
x(t) = -4e^t(e^t) + (1/3)e64t - (1/3)e^t

x(t) = -4e^2t + (1/3)e^4t - (1/3)e^t
y(t) = e^t

Thank you for your post. Your approach to solving the system of equations using Laplace transforms seems correct. However, there are a few minor errors in your calculations that I would like to point out.

Firstly, in the equation for Y(s), the term -4Y(s)/(s-1) should be -4Y(s)/(s+1). This is because the transform of e^4t is 1/(s-4), so when you substitute s=4, you should get 1/(4-4) = 1, not 0.

Secondly, in the calculation for the constants A, B, and C, you have used the wrong values for s. When substituting s=1, 3, and 4, you should get 1=A(1-1)(1-3) + B(4-4)(4-3) + C(4-4)(4-1), which simplifies to 1=0+0+C(0), so C=1. Similarly, when substituting s=1 and 3, you should get 1=A(1-1)(1-3) + B(3-4)(3-3) + C(3-4)(3-1), which simplifies to 1=0+B(0)+0, so B=1. Finally, when substituting s=1, you should get 1=A(1-1)(1-3) + B(1-4)(1-3) + C(1-4)(1-1), which simplifies to 1=0+0+C(0), so C=1. Therefore, the correct values for the constants are A=1/3, B=1/3, and C=1.

Finally, in your final answer for x(t), the term -4e^t(e^t) should be -4e^4t. This is because you have already substituted s=4 in the equation for X(s), so the term e^4t becomes 1/(4-4) = 1, not e^4t.

x(t) = -4e^4t + (1/3)e^4t - (1/3)e^t = (1/3)e^4t

## 1. How do I use Laplace to solve two equations?

To use Laplace to solve two equations, you will need to first transform the equations into their Laplace form. This involves taking the Laplace transform of each term in the equation and simplifying it. Then, you can use algebraic methods to solve for the variables in the equation.

## 2. What are the benefits of using Laplace to solve equations?

Laplace provides an efficient and systematic method for solving differential equations. It allows for the transformation of complex equations into simpler algebraic equations, making it easier to find solutions. It is also useful for solving equations with initial conditions or boundary conditions.

## 3. Can Laplace only be used to solve linear equations?

No, Laplace can also be used to solve nonlinear equations. However, the transformation process may be more complex and require more advanced techniques such as partial fraction decomposition.

## 4. What are some common mistakes when using Laplace to solve equations?

One common mistake is not properly applying the Laplace transform to each term in the equation. Another mistake is not simplifying the resulting algebraic equations correctly. It is also important to check for any initial or boundary conditions and include them in the final solution.

## 5. Are there any limitations to using Laplace to solve equations?

Laplace may not be suitable for all types of equations, especially those with discontinuous functions or non-constant coefficients. It also may not be the most efficient method for solving equations with multiple independent variables. In some cases, other methods such as numerical approximation may be more appropriate.

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