initial value problems using laplace tranforms

**mathprincess24** · April 4th 2009, 11:25 AM

i am stuck on these two problems. i can apply laplace tranforms and then i cant figure out how to solve for the inverse to finish the problem.

1) X” -4X = 3t ; X(0) = 0 ; X’(0) =0

1)X” + 4X’ +13X = te-t ; X(0)= 0 ; X’(0) = 2

please help!!

**mathprincess24** · April 4th 2009, 11:25 AM

in the second problem, e is raised to the power -t

**Chris L T521** · April 4th 2009, 11:58 AM

Originally Posted by mathprincess24

i am stuck on these two problems. i can apply laplace tranforms and then i cant figure out how to solve for the inverse to finish the problem.

1) X” -4X = 3t ; X(0) = 0 ; X’(0) =0

Applying the Laplace Transform, we see that $\mathcal{L}\left\lbrace x^{\prime\prime}\right\rbrace-4\mathcal{L}\!\left\lbrace x\right\rbrace=3\mathcal{L}\!\left\lbrace t\right\rbrace\implies s^2X\!\left(s\right)-s x\!\left(0\right)-x^{\prime}\!\left(0\right)-4X\!\left(s\right)=\frac{3}{s^2}$

Now, applying the initial equations, we are left with $s^2X\!\left(s\right)-4X\!\left(s\right)=\frac{3}{s^2}$

Solving for $X\!\left(s\right)$ , we have $X\!\left(s\right)=\frac{3}{s^2\left(s^2-4\right)}$

Take note that $s^2-(s^2-4)=4$

So, we have $X\!\left(s\right)=\frac{3}{4}\frac{4}{s^2\left(s^2-4\right)}=\frac{3}{4}\frac{s^2-\left(s^2-4\right)}{s^2\left(s^2-4\right)}=\frac{3}{4}\left[\frac{1}{s^2-4}-\frac{1}{s^2}\right]=\frac{3}{8}\left(\frac{2}{s^2-4}\right)-\frac{3}{4}\left(\frac{1}{s^2}\right)$

Taking the inverse Laplace Transform, we have $\frac{3}{8}\mathcal{L}^{-1}\!\left\lbrace\frac{2}{s^2-4}\right\rbrace-\frac{3}{4}\mathcal{L}^{-1}\!\left\lbrace\frac{1}{s^2}\right\rbrace=\boxed{ \frac{3}{8}\sinh\!\left(2t\right)-\frac{3}{4}t}$

Does this make sense?

1)X” + 4X’ +13X = te-t ; X(0)= 0 ; X’(0) = 2

please help!!

For this one, its somewhat similar to the first one. I'll get you started:

Apply the Laplace Transform to both sides, you should have

$\mathcal{L}\!\left\lbrace x^{\prime\prime}\right\rbrace+4\mathcal{L}\!\left\ lbrace x^{\prime}\right\rbrace+13\mathcal{L}\!\left\lbrac e x\right\rbrace$ $=\mathcal{L}\!\left\lbrace te^{-t}\right\rbrace\implies s^2X\!\left(s\right)-s x\!\left(0\right)-x^{\prime}\!\left(0\right)+4\left[sX\!\left(s\right)-x\!\left(0\right)\right]+13X\!\left(s\right)=\frac{1}{\left(s+1\right)^2}$

Applying the initial conditions, we're left with $s^2X\!\left(s\right)-2+4sX\!\left(s\right)+13X\!\left(s\right)=\frac{1} {\left(s+1\right)^2}$

Solving for $X\!\left(s\right)$ , we have $\left(s^2+4s+13\right)X\!\left(s\right)=\frac{1}{\ left(s+1\right)^2}+2$ $\implies X\!\left(s\right)=\frac{1}{\left(s^2+4s+13\right)\ left(s+1\right)^2}+\frac{2}{s^2+4s+13}$

Can you try to find the inverse Laplace transform on your own? For the first term, apply partial fractions. On the second, my hint to you is $s^2+4s+13=\left(s+2\right)^2+9$ (hinting at the fact that a translation is required)

**mathprincess24** · April 6th 2009, 09:41 AM

Everything that you did i understand. i can do all that on my own, but going about finding the inverse is what i get stuck on. i barely remember doing partial fractions so i'm a bit rusty.

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