# Thread: laplace transforms to solve IVPs

1. ## laplace transforms to solve IVPs

Use Laplace Transforms to solve the following Initial Value Problems:

(i)

$\displaystyle \dot{y_1} = y_{2}$
$\displaystyle \dot{y_2} = -y_{1} + e^t$

with $\displaystyle y_{1}(0) = 0, y_{2}(0) = 0$

(ii)

$\displaystyle \dddot{y} = \begin{cases} t & \ t < 3 \\ 0 & \ t \geq 3 \end{cases}$ with $\displaystyle y(0) = 0$, $\displaystyle \dot{y}(0) = 0$ and $\displaystyle \ddot{y}(0) = 0$

2. Originally Posted by wik_chick88
Use Laplace Transforms to solve the following Initial Value Problems:

(i)

$\displaystyle \dot{y_1} = y_{2}$
$\displaystyle \dot{y_2} = -y_{1} + e^t$

with $\displaystyle y_{1}(0) = 0, y_{2}(0) = 0$
There are two ways (at least) of doing this the first involves differentiating one of the equation and then substituting the other into the new equation:

$\displaystyle \ddot{y}_1=\dot{y}_2=-y_{1} + e^t$

Then solving the new equation for \dot y:

$\displaystyle \ddot{y}_1=-y_{1} + e^t$

Laplace transforming:

$\displaystyle s^2Y_1(s)-s y_1(0)-\dot{y}_1(0)=Y_1(s)-\frac{1}{s-1}, \ \ \ \ s>1$

Applying the initial conditions $\displaystyle y_1(0)=0$, $\displaystyle \dot{y}_1=0$ this becomes:

$\displaystyle Y_1(s)=-\frac{1}{(s-1)^2(s+1)}$

which you invert by the use of partial fractions and a table of inverse Laplace transforms.

The second method involves writing this as a vector ODE:

$\displaystyle \dot{ {\bold{y}} }={\bold{A}}{\bold{y}} + (0,e^t)^T$

where:

$\displaystyle {\bold{A}}=\left[\begin{array}{rr}0&1\\-1&0\end{array}\right]$

and: $\displaystyle {\bold{y}}=(y_1,y_2)^T$

Now take the Laplace transform:

$\displaystyle s{ {\bold{Y}} }(s)-{\bold{y}}(0)={\bold{A}}{\bold{Y}}(s) + (0,-1/(s-1))^T$

Then (as :$\displaystyle {\bold{y}}(0)=(0,0)^T$)

$\displaystyle {\bold{Y}}(s)=({\bold{A}}-s{\bold{I}})^{-1}(0,1/(s-1))^T$

Now with some matrix algebra you can find $\displaystyle Y_1(s)$ and $\displaystyle Y_2(s)$ and proceed as before.

CB

3. Originally Posted by wik_chick88

(ii)

$\displaystyle \dddot{y} = \begin{cases} t & \ t < 3 \\ 0 & \ t \geq 3 \end{cases}$ with $\displaystyle y(0) = 0$, $\displaystyle \dot{y}(0) = 0$ and $\displaystyle \ddot{y}(0) = 0$
Here you have:

$\displaystyle \dddot{y} =u(t-3) t$

where u(.) is the Heavyside unit step function, and:

$\displaystyle \mathcal{L}\dddot{y}=s^3Y(s)-s^2 \ddot{y}(0)-s\dot{y}(0)-y(0)$.

The rest is routine.

CB

4. ... use Laplace Transforms to solve the following Initial Value Problems:

$\displaystyle \dddot{y} = \begin{cases} t & \ t < 3 \\ 0 & \ t \geq 3 \end{cases}$ with $\displaystyle y(0) = 0$, $\displaystyle \dot{y}(0) = 0$ and $\displaystyle \ddot{y}(0) = 0$

If we intend to apply the LT for resolving a DE, then we have to impose that $\displaystyle t \ge 0$. So the proposed third order differrential equation must be written as…

$\displaystyle \dddot{y} = \begin{cases} t & \ 0 \le t < 3 \\ 0 & \ t \geq 3 \end{cases} =$ $\displaystyle t\cdot u(t) - (3+t) \cdot u(t-3)$

... where $\displaystyle u(*)$ is the Heavyside unit step function...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Originally Posted by chisigma
If we intend to apply the LT for resolving a DE, then we have to impose that $\displaystyle t \ge 0$. So the proposed third order differrential equation must be written as…
No we don't, the default LT in use is the one sided LT and so what the function does at negative arguments is irrelevant to the forward transform.

But the inverse transform is only valid for $\displaystyle t\ge 0$, since any function that agree on the non-negative half real axis have the same (one-sided) LT.

So the solution obtained using the LT to a IVP is only valid for non-negative times.

CB

6. Originally Posted by CaptainBlack
There are two ways (at least) of doing this the first involves differentiating one of the equation and then substituting the other into the new equation:

$\displaystyle \ddot{y}_1=\dot{y}_2=-y_{1} + e^t$

Then solving the new equation for \dot y:

$\displaystyle \ddot{y}_1=-y_{1} + e^t$

Laplace transforming:

$\displaystyle s^2Y_1(s)-s y_1(0)-\dot{y}_1(0)=Y_1(s)-\frac{1}{s-1}, \ \ \ \ s>1$

Applying the initial conditions $\displaystyle y_1(0)=0$, $\displaystyle \dot{y}_1=0$ this becomes:

$\displaystyle Y_1(s)=-\frac{1}{(s-1)^2(s+1)}$

which you invert by the use of partial fractions and a table of inverse Laplace transforms.
hi thanks so much! i do have a question though...
when you laplace transformed it, why did the $\displaystyle -y_{1} + e^t$ get laplace transformed to $\displaystyle Y_{1}(s) - \frac{1}{s-1}$ wouldn't it be $\displaystyle -Y_{1}(s) + \frac{1}{s-1}$??