Coupled Differential Equations

**Downes** · August 21st 2011, 02:29 AM

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

**Prove It** · August 21st 2011, 06:23 AM

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?

**Downes** · August 21st 2011, 06:27 AM

It asks to solve it

**Prove It** · August 21st 2011, 06:31 AM

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.

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