System of DE's.

• November 26th 2010, 05:47 PM
rafi98
System of DE's.
dx/dt + y = sint , x(0)=2
dy/dt + x = cost , y(0)=0

this is one problem that comes in my laplace transform chapter. Any method can be used
• November 26th 2010, 06:09 PM
Prove It
From DE 1: $\displaystyle \frac{dx}{dt} + y = \sin{t}$

$\displaystyle y = -\frac{dx}{dt} + \sin{t}$

$\displaystyle \frac{dy}{dt} = -\frac{d^2x}{dt^2} + \cos{t}$.

Substituting into DE 2:

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

$\displaystyle -\frac{d^2x}{dt^2} + x = 0$

$\displaystyle \frac{d^2x}{dt^2} - x = 0$.

Now, the characteristic equation is $\displaystyle m^2 - 1 = 0$

$\displaystyle m^2 = 1$

$\displaystyle m = \pm 1$.

So the solution to the homogeneous equation is $\displaystyle x = Ae^{t} + Be^{-t}$.

Follow a similar process to find $\displaystyle y$.
• November 26th 2010, 06:19 PM
rafi98
Quote:

Originally Posted by Prove It
From DE 1: $\displaystyle \frac{dx}{dt} + y = \sin{t}$

$\displaystyle y = -\frac{dx}{dt} + \sin{t}$

$\displaystyle \frac{dy}{dt} = -\frac{d^2x}{dt^2} + \cos{t}$.

Substituting into DE 2:

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

$\displaystyle -\frac{d^2x}{dt^2} + x = 0$

$\displaystyle \frac{d^2x}{dt^2} - x = 0$.

Now, the characteristic equation is $\displaystyle m^2 - 1 = 0$

$\displaystyle m^2 = 1$

$\displaystyle m = \pm 1$.

So the solution to the homogeneous equation is $\displaystyle x = Ae^{t} + Be^{-t}$.

Follow a similar process to find $\displaystyle y$.

But if we do like this , we may not be able to make use of the condition given in the question. Can i ?. Please respond. Thank you
• November 26th 2010, 06:26 PM
Prove It
Quote:

Originally Posted by rafi98
But if we do like this , we may not be able to make use of the condition given in the question. Can i ?. Please respond. Thank you

Find $\displaystyle y$ first, see what you get.
• November 26th 2010, 07:42 PM
rafi98
What is small letter a and b . And i dont know the formula of your second reply.
• November 26th 2010, 07:45 PM
Prove It
Quote:

Originally Posted by rafi98
dx/dt + y = sint , x(0)=2
dy/dt + x = cost , y(0)=0

this is one problem that comes in my laplace transform chapter. Any method can be used

From the second DE

$\displaystyle \frac{dy}{dt} + x = \cos{t}$

$\displaystyle x = -\frac{dy}{dt} + \cos{t}$

$\displaystyle \frac{dx}{dt} = -\frac{d^2y}{dt^2} - \sin{t}$.

Substitute into the first DE and solve for $\displaystyle y$.
• November 27th 2010, 04:32 AM
Krizalid
Quote:

Originally Posted by rafi98
dx/dt + y = sint , x(0)=2
dy/dt + x = cost , y(0)=0

this is one problem that comes in my laplace transform chapter. Any method can be used

apply the Laplace Transform and get

$s\widehat x(s)-x(0)+\widehat y(s)=\dfrac1{s^2+1}$ and $s\widehat y(s)-y(0)+\widehat x(s)=\dfrac s{s^2+1},$ the rest is algebra and a matter of finding inverse transforms.