# Math Help - ODE

1. ## 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..

2. Use:

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

http://www.fernandorevilla.es/

3. 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

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

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

5. 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.

,

with the condition,

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

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

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

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

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

Now stuck here..

6. Good.

7. Originally Posted by General
Good.
Ok, After simplifying, I got,

$\mathcal{L}(y)=\frac{3p^4+2p^3+4p^2+4p+1}{(p^3-1)(p^2+1)}$

Now what to do? I'm not getting any Idea to more simplify it..

8. Note $(p^3-1)=(p-1)(p^2+p+1)$

You will use partial fraction expansion.
But it will be too long x_x