# Math Help - Solving a second order DE using the method of Laplace Transforms

1. ## Solving a second order DE using the method of Laplace Transforms

can anyone help me to solve this problem [using laplace Transforms] ...just wanna know if we have the same answers..i have difficulty in answering the partial fraction of this prob...3/8 d^2x/dt^2 + 24x =24sin4t where: x(0)=1/3 and x'(0)= 0

2. Originally Posted by mhyann
can anyone help me to solve this problem [using laplace Transforms] ...just wanna know if we have the same answers..i have difficulty in answering the partial fraction of this prob...3/8 d^2x/dt^2 + 24x =24sin4t where: x(0)=1/3 and x'(0)= 0
If you have an answer I assume you've done the calculation. If you post your calculation and answer someone will be able to check it.

3. Originally Posted by mhyann
can anyone help me to solve this problem [using laplace Transforms] ...just wanna know if we have the same answers..i have difficulty in answering the partial fraction of this prob...3/8 d^2x/dt^2 + 24x =24sin4t where: x(0)=1/3 and x'(0)= 0

$\frac{3}{8}x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right)$ where $x\left( 0 \right) = \frac{1}{3}$ and $x'\left( 0 \right) = 0$.

First, simplifying the equation: $x''\left( t \right) + 64x\left( t \right) = 64\sin \left( {4t} \right)$.

$\sin \left( {4t} \right)\underset{\raise0.3em\hbox{\smash{\scripts criptstyle\cdot}}}{ \doteq } \frac{4}{{{s^2} + 16}}$

$x\left( t \right)\underset{\raise0.3em\hbox{\smash{\scripts criptstyle\cdot}}}{ \doteq } \overline {x\left( s \right)}$

$x''\left( t \right)\underset{\raise0.3em\hbox{\smash{\scripts criptstyle\cdot}}}{ \doteq } {s^2}\overline {x\left( s \right)} - x\left( 0 \right)s - x'\left( 0 \right) = {s^2}\overline {x\left( s \right)} - \frac{s}{3}$

So, we have

${s^2}\overline {x\left( s \right)} - \frac{s}{3} + 64\overline {x\left( s \right)} = \frac{{256}}{{{s^2} + 16}} \Rightarrow$

$\Rightarrow \left( {{s^2} + 64} \right)\overline {x\left( s \right)} = \frac{{256}}{{{s^2} + 16}} + \frac{s}{3} \Rightarrow$

$\Rightarrow \overline {x\left( s \right)} = \frac{s}{{3\left( {{s^2} + 64} \right)}} + \frac{{256}}{{\left( {{s^2} + 16} \right)\left( {{s^2} + 64} \right)}}$

$\frac{s}{{3\left( {{s^2} + 64} \right)}}\underset{\raise0.3em\hbox{\smash{\scrip tscriptstyle\cdot}}}{ \doteq } \frac{1}{3}\cos \left( {8t} \right)$

$\frac{{256}}{{\left( {{s^2} + 16} \right)\left( {{s^2} + 64} \right)}} = \frac{{16}}{{3\left( {{s^2} + 16} \right)}} - \frac{{16}}{{3\left( {{s^2} + 64} \right)}} =$

$= \frac{4}{3}\frac{4}{{{s^2} + 16}} - \frac{2}{3}\frac{8}{{{s^2} + 64}}\underset{\raise0.3em\hbox{\smash{\scriptscri ptstyle\cdot}}}{ \doteq } \frac{4}{3}\sin \left( {4t} \right) - \frac{2}{3}\sin \left( {8t} \right)$

$\overline {x\left( s \right)} {\underset{\raise0.3em\hbox{\smash{\scriptscripts tyle\cdot}}}{ \doteq } } \frac{1}{3}\cos \left( {8t} \right) + \frac{4}{3}\sin \left( {4t} \right) - \frac{2}{3}\sin \left( {8t} \right)$

Finally, we have

$\boxed{x\left( t \right) = \frac{1}{3}\cos \left( {8t} \right) + \frac{4}{3}\sin \left( {4t} \right) - \frac{2}{3}\sin \left( {8t} \right).}$

4. ## thanks for helping me

but its not 64 its 24sin4(t)..anyway i will follow the concept.

muchas gracias to u man...

5. Originally Posted by mhyann
but its not 64 its 24sin4(t)..anyway i will follow the concept.
In your first post you wrote is not clear:

thise $\frac{3}{8}x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right)$ or thise $x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right)$???

If it was thise $x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right)$, then your answer must be $x\left( t \right) = \frac{1}{3}\cos \left( {2\sqrt 6 t} \right) - \sqrt 6 \sin \left( {2\sqrt 6 t} \right) + 3\sin \left( {4t} \right)$.

6. ## the first one with 3/8

i get u now.u multiplied the 8 to 24 then divided by 3 that's why it comes up to 64.maybe you we're right....maybe i was wrong in distributing the term/numbers.but we almost have the same answer...hehehe ive got another mistakes in exam.i'll keep working on this promise...
thank u for helping me dude!