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
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
$\displaystyle \frac{3}{8}x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right)$ where $\displaystyle x\left( 0 \right) = \frac{1}{3}$ and $\displaystyle x'\left( 0 \right) = 0$.
First, simplifying the equation: $\displaystyle x''\left( t \right) + 64x\left( t \right) = 64\sin \left( {4t} \right)$.
$\displaystyle \sin \left( {4t} \right)\underset{\raise0.3em\hbox{$\smash{\scripts criptstyle\cdot}$}}{ \doteq } \frac{4}{{{s^2} + 16}}$
$\displaystyle x\left( t \right)\underset{\raise0.3em\hbox{$\smash{\scripts criptstyle\cdot}$}}{ \doteq } \overline {x\left( s \right)}$
$\displaystyle x''\left( t \right)\underset{\raise0.3em\hbox{$\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
$\displaystyle {s^2}\overline {x\left( s \right)} - \frac{s}{3} + 64\overline {x\left( s \right)} = \frac{{256}}{{{s^2} + 16}} \Rightarrow$
$\displaystyle \Rightarrow \left( {{s^2} + 64} \right)\overline {x\left( s \right)} = \frac{{256}}{{{s^2} + 16}} + \frac{s}{3} \Rightarrow$
$\displaystyle \Rightarrow \overline {x\left( s \right)} = \frac{s}{{3\left( {{s^2} + 64} \right)}} + \frac{{256}}{{\left( {{s^2} + 16} \right)\left( {{s^2} + 64} \right)}}$
$\displaystyle \frac{s}{{3\left( {{s^2} + 64} \right)}}\underset{\raise0.3em\hbox{$\smash{\scrip tscriptstyle\cdot}$}}{ \doteq } \frac{1}{3}\cos \left( {8t} \right)$
$\displaystyle \frac{{256}}{{\left( {{s^2} + 16} \right)\left( {{s^2} + 64} \right)}} = \frac{{16}}{{3\left( {{s^2} + 16} \right)}} - \frac{{16}}{{3\left( {{s^2} + 64} \right)}} =$
$\displaystyle = \frac{4}{3}\frac{4}{{{s^2} + 16}} - \frac{2}{3}\frac{8}{{{s^2} + 64}}\underset{\raise0.3em\hbox{$\smash{\scriptscri ptstyle\cdot}$}}{ \doteq } \frac{4}{3}\sin \left( {4t} \right) - \frac{2}{3}\sin \left( {8t} \right)$
$\displaystyle \overline {x\left( s \right)} {\underset{\raise0.3em\hbox{$\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
$\displaystyle \boxed{x\left( t \right) = \frac{1}{3}\cos \left( {8t} \right) + \frac{4}{3}\sin \left( {4t} \right) - \frac{2}{3}\sin \left( {8t} \right).}$
In your first post you wrote is not clear:
thise $\displaystyle \frac{3}{8}x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right)$ or thise $\displaystyle x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right)$???
If it was thise $\displaystyle x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right)$, then your answer must be $\displaystyle x\left( t \right) = \frac{1}{3}\cos \left( {2\sqrt 6 t} \right) - \sqrt 6 \sin \left( {2\sqrt 6 t} \right) + 3\sin \left( {4t} \right)$.
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!