1. ## 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

2. ## Re: Coupled Differential Equations

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?

4. ## Re: Coupled Differential Equations

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.