# Differential equation

Assume, that y is a differentiable function in x that satisfy the equation

$x^2-3xy+y^3=7$

Find y' and y'' in the point (x,y)=(4,3)

How can i find y' and y'' when it's not straightforward to isolate y?

• Try to derive the whole equation with respect to $x$ – popoolmica Oct 4 '15 at 17:23
• Differentiate with respect to $x$ and substitute values. You can factor out $y'(x)$ in order to compute a value.Then repeat. – copper.hat Oct 4 '15 at 17:23
• Search this site for "implicit differentiation" and you will get lots of examples where such questions are treated. See for example here or here. – mickep Oct 4 '15 at 17:26
• Implicit differentiation is the key. Just remember $y$ is a function of $x$ so you'll need the product rule and chain rule. – Karl Oct 4 '15 at 17:26

$$x^2-3xy+y^3=7$$

Differentiate w.r.t. $x$ both sides

$2x -3y -3xy' +3y^2y' = 0 \tag1$

Differentiate w.r.t. $x$ again

$2-3y'-3y' -3xy'' +6y(y')^2 +3y^2y'' = 0\tag2$

Put $x=4,y=3$ in $(1)$

$8-9-12y' + 27y' =0 \implies y' =\frac{1}{15}$

Put $x=4,y=3,y'=\frac{1}{15}$ in $(2)$

$2-\frac{1}{5}-\frac{1}{5}-12y'' +\frac{18}{225} +27y'' =0 \implies y'' =-\frac{378}{3375}$

• I don't get how you get -3y and $+2y^2y'$ in (1) I get the same as H.R above. – Christian Skjøth Oct 4 '15 at 18:21
• $x$ is not a constant. so we have to use the product rule to differentiate. – user265328 Oct 4 '15 at 18:23
• But isn't $y^3=y^2y$ so using product rule we get $2yy'\cdot y+y^2y'=3y^2y'$ – Christian Skjøth Oct 4 '15 at 21:53
• oops, my bad. i'll correct that. – user265328 Oct 5 '15 at 2:22

You have

$${x^2} - 3xy + {y^3} = 7$$

Consider that $y=y(x)$ and then differentiate this equation to obtain

$$2x - 3xy' + 3{y^2}y' = 0$$

now, solve for ${y'}$ to get

$$y' = {{2x} \over {3x - 3{y^2}}}$$

Now, put $(4,3)$ into the above equation to get

$$y'(4) = {8 \over {12 - 27}} = {8 \over { - 15}}$$

You can do the same things to obtain ${y''}$ at 4.