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

Printable View

- Aug 21st 2011, 02:29 AMDownesCoupled 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 - Aug 21st 2011, 06:23 AMProve ItRe: Coupled Differential Equations
- Aug 21st 2011, 06:27 AMDownesRe: Coupled Differential Equations
It asks to solve it

- Aug 21st 2011, 06:31 AMProve ItRe: Coupled Differential Equations
From the first equation we have $\displaystyle \displaystyle \frac{dy}{dx} + z = x \implies z = x - \frac{dy}{dx} \implies \frac{dz}{dx} = 1 - \frac{d^2y}{dx^2}$.

Substituting this into the second equation gives

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

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