# ODE

• December 26th 2010, 12:25 AM
kjchauhan
ODE
Please solve the following ODE using Laplace Transform::

$\frac{d^3y}{dt^3} - \frac{dy}{dt} = 2\cos t$,

with the condition, $y(0)=3, y'(0)=2, y''(0)=1$

I've to use: $\mathcal{L}({y'''})=p^3\mathcal{L}({y})-p^2y(0)-py'(0)-y''(0)$

But I couldn't get it.. So please give me solution..

• December 26th 2010, 12:37 AM
FernandoRevilla
Use:

$\mathcal{L}\{2\cos t\}=\dfrac{2p}{p^2+1}$

http://www.fernandorevilla.es/
• December 26th 2010, 12:41 AM
Prove It
You don't need to, if you make the substitution $\displaystyle Y = \frac{dy}{dt}$ this transforms the DE into

$\displaystyle \frac{d^2Y}{dt^2} - Y = 2\cos{t}$

which is second order linear constant coefficient nonhomogeneous.

Homogeneous solution:

The characteristic equation is

$\displaystyle m^2 - 1 = 0$

$\displaystyle m^2 = 1$

$\displaystyle m = \pm 1$.

So the homogeneous solution is $\displaystyle Y = C_1e^t + C_2e^{-t}$.

Nonhomogeneous solution:

Assume a solution of the form $\displaystyle Y = A\sin{t} + B\cos{t}$.

Then $\displaystyle \frac{dY}{dt} = A\cos{t} - B\sin{t}$ and $\displaystyle \frac{d^2y}{dt^2} = -A\sin{t} - B\cos{t}$.

Substituting into the DE $\displaystyle \frac{d^2Y}{dt^2} - Y = 2\cos{t}$ gives

$\displaystyle -A\sin{t} - B\cos{t} - (A\sin{t} - B\cos{t}) = 2\cos{t}$

$\displaystyle -2A\sin{t} - 2B\cos{t} = 0\sin{t} + 2\cos{t}$

$\displaystyle -2A = 0$ and $\displaystyle -2B = 2$

$\displaystyle A = 0$ and $\displaystyle B = -1$.

So the nonhomogeneous solution is $\displaystyle Y = -\cos{t}$.

The general solution is the sum of the homogeneous and nonhomogeneous solutions.

So $\displaystyle Y = C_1e^{t} + C_2e^{-t} - \cos{t}$.

From this we know $\displaystyle \frac{dY}{dt} = C_1e^{t} - C_2e^{-t} - \cos{t}$.

From the initial conditions:

$\displaystyle C_1e^0 + C_2e^{-0} - \cos{0} = 2$ and $\displaystyle C_1e^0 - C_2e^{-0} - \cos{0} = 1$

$\displaystyle C_1 + C_2 - 1 = 2$ and $\displaystyle C_1 - C_2 - 1 = 1$

$\displaystyle C_1 + C_2 = 3$ and $\displaystyle C_1 - C_2 = 2$

$\displaystyle (C_1 + C_2) + (C_1 - C_2) = 3 + 2$

$\displaystyle 2C_1 = 5$

$\displaystyle C_1 = \frac{5}{2}$.

So $\displaystyle C_2 = \frac{1}{2}$.

Therefore $\displaystyle Y = \frac{5}{2}e^t + \frac{1}{2}e^{-t} - \cos{t}$

$\displaystyle \frac{dy}{dt} = \frac{5}{2}e^t + \frac{1}{2}e^{-t} - \cos{t}$

$\displaystyle y = \int{\frac{5}{2}e^t + \frac{1}{2}e^{-t} - \cos{t}\,dt}$

$\displaystyle y = \frac{5}{2}e^t - \frac{1}{2}e^{-t} - \sin{t} + C$.

From the final initial condition

$\displaystyle 3 = \frac{5}{2}e^0 - \frac{1}{2}e^{-0} - \sin{0} + C$

$\displaystyle 3 = \frac{5}{2} - \frac{1}{2} - 0 + C$

$\displaystyle 3 = 2 + C$

$\displaystyle C=1$.

So finally our final solution is

$\displaystyle y = \frac{5}{2}e^t - \frac{1}{2}e^{-t} - \sin{t} + 1$.
• December 26th 2010, 01:42 AM
General
We do not give solutions here.
Your first step is taking the laplace of both sides.
Do it and tell us if you stuck.
• December 26th 2010, 07:25 AM
kjchauhan
Quote:

Originally Posted by General
We do not give solutions here.
Your first step is taking the laplace of both sides.
Do it and tell us if you stuck.

http://www.mathhelpforum.com/math-he...e5956950bc.png,

with the condition, http://www.mathhelpforum.com/math-he...795ce6ab0f.png

$\therefore \mathcal{L}(y''') - \mathcal{L}(y) = 2\mathcal{L}(\cos t)$

$\therefore p^3\mathcal{L}(y)-p^2y(0)-py'(0)-y''(0)- \mathcal{L}(y)=\frac{2p}{p^2+1}$

$\therefore p^3\mathcal{L}(y)-3p^2-2p-1-\mathcal{L}(y)=\frac{2p}{p^2+1}$

$\therefore (p^3-1)\mathcal{L}(y)=3p^2+2p+1+\frac{2p}{p^2+1}$

$\therefore \mathcal{L}(y)=\frac{3p^2+2p+1+\frac{2p}{p^2+1}}{p ^3-1}$

Now stuck here..
• December 26th 2010, 07:31 AM
General
Good.
• December 26th 2010, 08:30 AM
kjchauhan
Quote:

Originally Posted by General
Good.
$\mathcal{L}(y)=\frac{3p^4+2p^3+4p^2+4p+1}{(p^3-1)(p^2+1)}$
Note $(p^3-1)=(p-1)(p^2+p+1)$