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Question: Determine whether the given relation is an implicit solution to the give differential equation. Assume that the relationship does define y implicitly as a function of x and use implicit differentiation.

13: $sin(y) + xy - x^3=2, y'' = \frac{6xy'+(y')^3sin(y)-2(y')^2}{3x^2-y}$

I am able to get the first derivative of $y'=\frac{3x^2-y}{cos(y)+x}$ but I am unsure what to do after that point. I tried to use the quotient rule, but I did not get the proper answer. The back of the book says that the equation is an implicit solution, but I am not getting that answer.

With the quotient rule I get. $y''=\frac{(3x^2-y)(sin(y)y'(x)-1)+(6x-y'(x))(x+cos(y))}{(x+cos(y))^2}$, which is not the same even after I simplified. I don't even know how they got the $3x^2-y$ in the denominator.

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It's best to avoid the quotient rule. Once you have implicitly differentiated once, differentiate again without rearranging. Then rearrange to get $y''$:

Differentiating once, $$ \cos y y'+xy'+y-3x^2=0$$ Differentiating again, $$-\sin y(y')^2+\cos y''+y'+xy''+y'-6x=0$$ $$\Rightarrow y''(\cos y+x)=6x-2y'+\sin y(y')^2$$ Now multiply both sides by $y'$ and replace $(\cos y y'+xy')$ with $(3x^2-y)$ and divide, and there you have it.

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  • $\begingroup$ Ok lets say that I have $y'=-3x^2+y'(x)(x+cos(y))+y$. Now I want to differentiate once again so lets say $g=dy/dx$ we have $y'=\frac{d}{dx}(-3x^2)+g'(x+cos(y))+g(\frac{d}{dx}(x+cos(y))+\frac{d}{dx}(y)$ $\endgroup$ – JuliusDariusBelosarius Aug 31 '15 at 13:24
  • $\begingroup$ I have added to my answer $\endgroup$ – David Quinn Aug 31 '15 at 14:57

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