Coupled Differential Equations

Printable View

August 21st 2011, 02:29 AM
Downes

Coupled Differential Equations

Hey, I've been set a question which i have no idea how to attempt. Please could some one show me some guidance? thanks

Question:

y'(x) +z(x)= x
z'(x) +4y(x)=0
y(0)=1
z(0)=-1
August 21st 2011, 06:23 AM
Prove It

Re: Coupled Differential Equations

Quote:

Originally Posted by Downes

Hey, I've been set a question which i have no idea how to attempt. Please could some one show me some guidance? thanks

Question:

y'(x) +z(x)= x
z'(x) +4y(x)=0
y(0)=1
z(0)=-1

What has the question asked you to do?
August 21st 2011, 06:27 AM
Downes

Re: Coupled Differential Equations

It asks to solve it
August 21st 2011, 06:31 AM
Prove It

Re: Coupled Differential Equations

Quote:

Originally Posted by Downes

Hey, I've been set a question which i have no idea how to attempt. Please could some one show me some guidance? thanks

Question:

y'(x) +z(x)= x
z'(x) +4y(x)=0
y(0)=1
z(0)=-1

From the first equation we have $\displaystyle \frac{dy}{dx} + z = x \implies z = x - \frac{dy}{dx} \implies \frac{dz}{dx} = 1 - \frac{d^2y}{dx^2}$ .

Substituting this into the second equation gives

$\displaystyle 1 - \frac{d^2y}{dx^2} + 4y = 0 \implies \frac{d^2y}{dx^2} - 4y = 1$ .

This is a second order linear constant coefficient nonhomogeneous DE, which is easy to solve.

All times are GMT -8. The time now is 04:56 PM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.