# Thread: Find y'' given x + xy + y =2

1. ## Find y'' given x + xy + y =2

I'm learning Calculus on my own this summer and I've encountered a simple problem which i can't seem to get the right answer to. Here is the problem and my solution:

Find y'' given x + xy + y =2
first derivative
1 + y + x(y') + y' = 0 took derivative of each term
1 + y + y'(x+1) = 0 simplified a bit
(-1-y)/(x+1) = y' simple algebra.
second derivative
[-y'(x+1) - 1(-1-y)]/[(x+1)^2] quotient rule
[-y'(x+1)]/[(x+1)^2] - (-1-y)/[(x+1)^2] = 0 restated in a more useful manner.
[-y'(x+1)]/[(x+1)^2] = (-1-y)/[(x+1)^2] simple algebra
-y'(x+1) = -1-y simple algebra
y'(x+1) = 1+y multiplied both sides by -1
y' = (1+y) / (x+1)

The answer the books gives is: [2(1+y)]/[(1=x)^2]

2. ## Re: Basic calculus problem.

Originally Posted by RussianScience

Find y'' given x + xy + y =2
Have a go this way,

$\displaystyle x + xy + y =2$

$\displaystyle xy + y =2-x$

$\displaystyle y(x + 1) =2-x$

$\displaystyle y =\frac{2-x}{x+1}$

Now find y' and y''

3. ## Re: Basic calculus problem.

Originally Posted by RussianScience
[-y'(x+1)]/[(x+1)^2] - (-1-y)/[(x+1)^2] = 0 restated in a more useful manner.
Very bad. You have just decreed that the second derivative is everywhere zero (0).

Try = y" rather than = 0.

4. ## Re: Basic calculus problem.

Originally Posted by TKHunny
Very bad. You have just decreed that the second derivative is everywhere zero (0).

Try = y" rather than = 0.
That clears so much up, thanks for finding my stupid mistake!

5. ## Re: Basic calculus problem.

An alternative (implicit differentiation):

$\displaystyle x+xy+y=2\Rightarrow 1+y+xy'+y'=0\Rightarrow y'+y'+xy''+y''=0\Rightarrow$

$\displaystyle y''=\dfrac{-2y'}{x+1}$

Of course, in this case we can substitute $\displaystyle y'=\ldots$ and $\displaystyle y=\ldots$ .