# Solving linear eqns using laplace transforms

1. Aug 10, 2008

### :|max

Hey can anyone tell me the method to solve this, I can do inverse linear eqns and such but i dont know how to solve this type of question:

Use Laplace transforms to solve the following systems of linear equations
(a)dx/dt= 3x(t) − 4y(t),
dy/dt= 2x(t) + 3y(t),
with initial conditions x(0) = 1, y(0) = 0.
(b)3 dx/dt+ 2 dy/dt= t,
2 dx/dt −dy/dt= 2t^2,
with initial conditions x(0) = 1, y(0) = 0.

thanks

2. Aug 10, 2008

Take the Laplace transform of each equation. There is a property of the transform that tells you how both $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$ relate to $$x(t), y(t)$$ and $$x(0), y(0)$$. This reduces the original system of differential equations to a system of linear equations in $$\mathcal{L}x(t)$$ and $$\mathcal{L}y(t)$$. Solve the system for these two unknowns, then use inverse Laplace transforms to find the original functions. (This is the quick, condensed version of the solution process).

3. Aug 11, 2008

### :|max

Do you mean like:

s(Lx(t)) - x(0) = 3(L(x(t)) - 4(Ly(t))

s(Ly(t)) - y(0) = 2(Lx(t)) + 3(Ly(t))

4. Aug 11, 2008

\begin{align*} s(\mathcal{L}x(t)) - x(0) & = 3(\mathcal{L}(x(t)) - 4(\mathcal{L}y(t))\\ s(\mathcal{L}y(t)) - y(0) & = 2(\mathcal{L}x(t)) + 3(\mathcal{L}y(t)) \end{align*}

is what I meant. Your initial conditions give values for $$x(0) \text{ and } y(0)$$, so plug those in. Your next step is to consider $$\mathcal{L}x(t)$$ and $$\mathcal{L}y(t)$$ as your unknowns. You have a system of 2 linear equations in two unknowns for them - solve it any way you like. You should (will, I have faith in you) come to a point where you have

\begin{align*} \mathcal{L}x(t) & = \text{ an expression with numbers and } s\\ \mathcal{L}y(t) & = \text{ an expression with numbers and } s \end{align*}

If you find the inverse Laplace transforms of the expressions on the right you should have the solutions to the original system of DEs. (Check them to make sure you haven't made an arithmetic error.)
Good luck.

5. Aug 11, 2008

### :|max

OK i had a quick go of it, couldn't get it out but I think I understand the method, just some calculation mistake, i got irrational roots...which i suppose could have actually been right. I will have a better go of it tomorrow, thanks for your help :)

6. Aug 11, 2008

Again, I haven't checked the details that led to the post I commented on above: IF those terms are correct (no judgement here at all) think about this:

\begin{align*} s(\mathcal{L}x(t)) - x(0) & = 3(\mathcal{L}(x(t)) - 4(\mathcal{L}y(t))\\ s(\mathcal{L}y(t)) - y(0) & = 2(\mathcal{L}x(t)) + 3(\mathcal{L}y(t))\\ \intertext{Use the initial conditions.} s\mathcal{L}x(t) - 1 & = 3\mathcal{L}x(t) - 4\mathcal{L}y(t)\\ s\mathcal{L}y(t) - 0 & = 2\mathcal{L}x(t)+3\mathcal{L}y(t)\\ \intertext{Rearrange} (s-3)\mathcal{L}x(t)+4\mathcal{L}y(t) & = 1\\ -2\mathcal{L}x(t)+(s-3)\mathcal{L}y(t)&=0 \end{align*}

Solve these equations for $$\mathcal{L}x(t) \text{ and } \mathcal{L}y(t)$$. The answers will be the $$\textbf{Laplace transforms}$$ of the functions you seek. You should be able to find, or look up in one of the tables in your text, the appropriate functions.
Even though it has been MANY years since I first encountered this type of problem, I do remember that after working through the first few the ideas needed for organizing my work began to come naturally. I hope you will soon say the same.

7. Aug 12, 2008

### Alex6200

-Take the Laplace transform of both equations. REMEMBER that x and y are functions, not variables (a common mistake I've seen is treating the function like a variable and transforming it to 1/s, don't do it)

-Seeing that the Laplace transform of X'(t) is the same as sX(t) - X(0). If that's hard for you to see, trying comparing the Laplace transform of a sine curve and a cosine curve, and looking at how multiplying by s is the same as differentiating on the time domain. At least for me, this took some time to get used to.

-Reduce the matrix to reduced row echelon form, and find the values of X(s) and Y(s). Then do an inverse laplace transform on X(s) and Y(s) to get X(t) and Y(t), which is your solution.

Good luck :)

8. Aug 12, 2008

### :|max

Hey thanks everyone, I've finally got it out, woo. Thanks for your help.