Basic Laplace Transform of a 1st order Differential Equation

**craig** · October 12th 2011, 01:19 PM

Just wondering if someone could have a quick look over what I've done so far:

Assuming zero initial conditions, solve the following DE:

$\dot{x} + x = 2$

The Laplace Transform gives:

$sX(s) - x_0 + \frac{1}{s^2} = \frac{2}{s}$

Rearranging and using the fact that $x_0 =0$ gives us

$X(s) = \frac{2}{s^2} - \frac{1}{s^3}$

$X(s) = 2\frac{1!}{s^{1+1}} - \frac{1}{2} \frac{2!}{s^{2+1}}$

So $x(s) = 2t - \frac{t^2}{2}$ ?

**Danny** · October 12th 2011, 02:09 PM

After taking the Laplace transform, where did the $\frac{1}{s^2}$ come from?

**craig** · October 12th 2011, 11:11 PM

Originally Posted by Danny

After taking the Laplace transform, where did the $\frac{1}{s^2}$ come from?

I took the Laplace transform of $x$ , is this not right?

**FernandoRevilla** · October 12th 2011, 11:40 PM

Originally Posted by craig

I took the Laplace transform of $x$ , is this not right?

$x$ is a dependent variable ( $x=x(t)$ ), so you'll obtain

$s\mathcal{L}\{x\}-x_0+\mathcal{L}\{x\}=\frac{2}{s}\Rightarrow (s+1)\mathcal{L}\{x\}=x_0+\frac{2}{s}\Rightarrow \ldots$

**craig** · October 13th 2011, 12:30 AM

Originally Posted by FernandoRevilla

$x$ is a dependent variable ( $x=x(t)$ ), so you'll obtain

$s\mathcal{L}\{x\}-x_0+\mathcal{L}\{x\}=\frac{2}{s}\Rightarrow (s+1)\mathcal{L}\{x\}=x_0+\frac{2}{s}\Rightarrow \ldots$

Thanks for the reply!

So given that $x_0 = 0$ , does that mean that we just have $\mathcal{L}\{x\}=\frac{2}{s(s+1)}$ .

Which I presume we express in partial fractions, and then solve for $x(t)$ ?

**craig** · October 13th 2011, 12:38 AM

So for $\mathcal{L}\{x\}=\frac{2}{s(s+1)} = \frac{2}{s} - \frac{2}{s+1}$ we have:

$\mathcal{L}\{x\}= 2\frac{1}{s} - 2\frac{1}{s-(-1)}$

So $x(t) = 2(1) - 2 (e^{-1t}) = 2(1-e^{-t)$ ?

Thanks again for the replies.

**FernandoRevilla** · October 13th 2011, 01:25 AM

Originally Posted by craig

So $x(t) = 2(1) - 2 (e^{-1t}) = 2(1-e^{-t)$ ?

Right.

**Ackbeet** · October 13th 2011, 01:43 AM

Originally Posted by craig

So for $\mathcal{L}\{x\}=\frac{2}{s(s+1)} = \frac{2}{s} - \frac{2}{s+1}$ we have:

$\mathcal{L}\{x\}= 2\frac{1}{s} - 2\frac{1}{s-(-1)}$

So $x(t) = 2(1) - 2 (e^{-1t}) = 2(1-e^{-t)$ ?

Thanks again for the replies.

Technically, since you're working with the one-sided LT (I assume), you have unit step functions multiplying everything. That is,

$x(t)=2u(t)(1-e^{-t}),$

and the solution is valid for non-negative t's.

**FernandoRevilla** · October 13th 2011, 02:05 AM

Originally Posted by Ackbeet

and the solution is valid for non-negative t's.

Also considering the theorem about the necessary form of the solutions for the equation $x^{(n)}+a_{n-1}x^{(n-1)}+\ldots +a_1x'+a_0x=b(t)$ .

**craig** · October 13th 2011, 02:18 AM

Originally Posted by FernandoRevilla

Also considering the theorem about the necessary form of the solutions for the equation $x^{(n)}+a_{n-1}x^{(n-1)}+\ldots +a_1x'+a_0x=b(t)$ .

Sorry would you mind explaining what you meant by this? I understood the first part about the Unit-Step function, just not sure what you meant here?

Thankyou again!

**FernandoRevilla** · October 13th 2011, 04:08 AM

Originally Posted by craig

Sorry would you mind explaining what you meant by this?

The Laplace transform of $x:[0,+\infty)\to\mathbb{R}$ is defined by $\mathcal{L}\{x(t)\}=\int_0^{+\infty}e^{-st}x(t)\;dt$ , so a priori we find solutions valid for $t\geq 0$ . Your equation has (by a well known theorem) a determined form for its unique solution $x(t)=k+Ce^{at}$ valid for all $t\in\mathbb{R}$ so, necessarily is valid the solution found by LT for all $t\in\mathbb{R}$ .

**Ackbeet** · October 13th 2011, 04:32 AM

Originally Posted by FernandoRevilla

The Laplace transform of $x:[0,+\infty)\to\mathbb{R}$ is defined by $\mathcal{L}\{x(t)\}=\int_0^{+\infty}e^{-st}x(t)\;dt$ , so a priori we find solutions valid for $t\geq 0$ . Your equation has (by a well known theorem) a determined form for its unique solution $x(t)=k+Ce^{at}$ valid for all $t\in\mathbb{R}$ so, necessarily is valid the solution found by LT for all $t\in\mathbb{R}$ .

Agreed. You could simply say that you extend the solution found by the LT method to the entire real line. The result of the inverse LT does have the unit step functions in it, and, in fact, does not satisfy the DE for negative t's. However, with the theorem you have invoked, you can extend the solution by eliminating the unit step function.

**craig** · October 13th 2011, 11:57 PM

Thankyou both! I don't think we've covered this in lectures yet but I think it makes sense.

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