# Thread: Find a Particular Solution, given initial condition

1. ## Find a Particular Solution, given initial condition

(x+y)dx+(3x+3y-4)dy=o ---> y(1)=1

x+y=u
dy=du-dx

3u-4=3x+3y-4

udx+ (3u-u)(du-dx)=0

Solving, we arrive at

dx + (3u-4)/(4-2u)=0

x- 1/2(2 ln(u-2)+ 3u)=c

2x-(2 ln (u-2) + 3u)=c

Substituting

2x-2ln(x+y-2) - 3 (x+y)=c
-x-3y-2ln(x+y-2)=c

x+3y+2ln(x+y-2)=c

So here are the problems I ran into:

The answer in the book has the following answer:
x+3y+2ln(2-x-y)=c

If you plug in the Initial Value for x and y, you get

1+3+ 2Ln(2-1-1)= 4+2Ln(0), which obviously creates a problem since LN(0) if undefined.

2. Let's continue a bit with your solution. So

$\ln 2 - x - y = \dfrac{c - x - 3y}{2}$
and exponentiating gives

$2 - x- y = e^{\frac{c-x-3y}{2}} = k e^{\frac{-x-3y}{2}}$.

Your IC of $y(1) = 1$ gives $k = 0$ and your solution is $y = 2 - x.$

3. The answer in the book says that the general solution is

x+3y+2 ln(2-x-y)=c.. the LN is taken of the whole expression in the parentheses. Where did the x and 3y go in your solution?

So we're solving for c, not for y.

The end result should be x+3y+2ln(2-x-y)=equals some value.

Thanks for your help!

4. The two answers

$x+3y+2 ln(2-x-y)=c$

$2 - x - y = k e^{\dfrac{- x - 3y}{2}}$

are equivalent however yours cannot recover the initial condition where as mine can.

This problem is similar to finding the solutions to

$\dfrac{dy}{dx} = y(1-y)$ subject to (i) $y(0) = 0$, and (ii) $y(0) = 1$.