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kpg0001 · Jan 31, 2011, 04:40 PM

Determine whether the given relation is an implicit solution to the given equation.

sin(y)+xy-x^3=2-----relation

y''=(6xy'+((y')^3)sin(y)-2(y')^2)/((3x^2)-y)-----equation

The book says yes it is a solution but for some reason I am not getting it. You have to use implicit differentiation(duh).

What I did, which should be the right steps, is get the second derivative and solve for it(go from y-y'-y'' and isolate y''). Maybe I'm missing something.

This is what I got:

y''=(2y'sin(y)-2y'+6x)/(cos(y)+x)

jcaron2 · Jan 31, 2011, 10:14 PM

Using implicit differentiation, we should be able to find y' in terms of x and y:

$\sin{(y)}+xy-x^3=2$

$D\left \{ \sin{(y)}+xy-x^3 \right \}=D\left \{ 2 \right \}$

$D\left \{ \sin{(y)} \right \} + D\left \{ xy \right \}-D\left \{ x^3 \right \}=D\left \{ 2 \right \}$

${y}' \cos{(y)} + x{y}' + y -3x^2 =0$

${y}' = \frac{3x^2-y}{\cos{(y)} + x}$

And differentiating a second time to find y'':

$D\left \{{y}' \right \}= D\left \{\frac{3x^2-y}{\cos{(y)} + x}\right \}$

${y}''=\frac{(6x-y')(\cos{(y)}+x)-(3x^2-y)({y}'\sin{(y)}+1)}{(\cos{(y)}+x)^2}$

Now we can go back to the equation for y' and rearrange a little:

$\cos{(y)} + x= \frac{3x^2-y}{{y}'}$

Plugging that back into the equation for y'', we get

${y}''=\frac{(6x-y') \cdot {y}'}{3x^2-y}-\frac{({y}'\sin{(y)}+1) \cdot ({y}')^2}{3x^2-y}=\frac{6x{y}'+({y}')^3\sin{(y)}-2({y}')^2}{3x^2-y}$

So I agree with your book. It definitely IS an implicit solution. Did that all make sense?