# Coupled Differential Equations

• August 21st 2011, 03:29 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, 07:23 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, 07:27 AM
Downes
Re: Coupled Differential Equations
• August 21st 2011, 07:31 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.