# Differential Equations: System of equations

## Homework Statement

I have the following 2 equations dx/dt = 3x+y and dy/dt = sqrt(y)
I need to find the general solution to the system.

## Homework Equations

dx/dt = 3x+y and dy/dt = sqrt(y)

## The Attempt at a Solution

i did the y equation first using separation of variables and got
2sqrt(y) = t+c
so... y = (2t+A)^2 where A = 2c

then i plugged this into the 2nd equation for y so i have dx/dt = 3x + (2t+2t)^2
i'm not sure how to solve this. I tried using method of undetermined coefficients but i keep getting stuck. Can someone please show me step by step how to solve this? Thank you

tiny-tim
Homework Helper
Hi SpiffyEh!

(have a square-root: √ )
2sqrt(y) = t+c
so... y = (2t+A)^2 where A = 2c

No, (t/2 + A).
then i plugged this into the 2nd equation for y so i have dx/dt = 3x + (2t+2t)^2
i'm not sure how to solve this. I tried using method of undetermined coefficients but i keep getting stuck.

(You need both a complementary solution and a particular solution … I assume you've got the former?)

Substituting u = t/2 + A may make it easier.

If it doesn't, show us how far you get, and where you're stuck, and then we'll know how to help!

oops sorry completely missed the t/2...

well when i start the method of undetermined coefficients i can ge the yh but the yp is where i have issues. Since it has to be something like the right hand side i can't figure out what to set it to. Is this the wrong method to use in this case?

tiny-tim
Homework Helper
Hi SpiffyEh!

Whenever I see a polynomial on the RHS, I always try a polynomial as the particular solution.

i understand that, i just don't see how it works out because of the c being in two different parts of the equation. I don't know if im just being stupid and thinking of it wrong or not.

tiny-tim