Find \(y''\) if \(y=y^3+xy+x^3\text{.}\)
Solution This problem concerns some function \(y(x)\) that is not given to us explicitly. All that we are told is that \(y(x)\) satisfies
\begin{equation*}
y(x)=y(x)^3+xy(x)+x^3
\tag{E1}
\end{equation*}
for all
\(x\text{.}\) We are asked to find
\(y''(x)\text{.}\) We cannot solve this equation to get an explicit formula for
\(y(x)\text{.}\) So we use implicit differentiation, as we did in Example
2.11.1. That is, we apply
\(\diff{}{x}\) to both sides of (E1). This gives
\begin{equation*}
y'(x)=3y(x)^2\,y'(x)+y(x)+x\,y'(x)+3x^2
\tag{E2}
\end{equation*}
which we can solve for \(y'(x)\text{,}\) by moving all \(y'(x)\)’s to the left hand side, giving
\begin{equation*}
\big[1-x-3y(x)^2\big]y'(x) = y(x)+3x^2
\end{equation*}
and then dividing across.
\begin{equation*}
y'(x) = \frac{y(x)+3x^2}{1-x-3y(x)^2}
\tag{E3}
\end{equation*}
To get \(y''(x)\text{,}\) we have two options.
Method 1. Apply \(\diff{}{x}\) to both sides of (E2). This gives
\begin{equation*}
y''(x)=3y(x)^2\,y''(x)+6y(x)\,y'(x)^2+2y'(x)+x\,y''(x)+6x
\end{equation*}
We can now solve for \(y''(x)\text{,}\) giving
\begin{equation*}
y''(x) = \frac{6x+2y'(x)+6y(x)y'(x)^2}{1-x-3y(x)^2}
\tag{E4}
\end{equation*}
Then we can substitute in (E3), giving
\begin{align*}
\amp y''(x) = 2\frac{3x+ \frac{y(x)+3x^2}{1-x-3y(x)^2}
+3y(x) \big(\frac{y(x)+3x^2}{1-x-3y(x)^2}\big)^2}
{1-x-3y(x)^2}\\
&= 2\frac{3x{[1\!-\!x\!-\!3y(x)^2]}^2+ [y(x)\!+\!3x^2][1\!-\!x\!-\!3y(x)^2]
+3y(x) {[y(x)\!+\!3x^2]}^2}{{[1-x-3y(x)^2]}^3}
\end{align*}
Method 2. Alternatively, we can also differentiate (E3).
\begin{align*}
\amp y''(x) = \frac{[y'(x)+6x][1\!-\!x\!-\!3y(x)^2]-
[y(x)+3x^2][-1-6y(x)y'(x)]}{{[1-x-3y(x)^2]}^2}\\
&= \frac{\big[\frac{y(x)+3x^2}{1-x-3y(x)^2}+6x\big][1-x-3y(x)^2]-
[y(x)+3x^2][-1-6y(x)\frac{y(x)+3x^2}{1-x-3y(x)^2}]}
{{[1-x-3y(x)^2]}^2}\\
&= \frac{2[y(x)+3x^2][1\!-\!x\!-\!3y(x)^2]+6x{[1\!-\!x\!-\!3y(x)^2]}^2
+6y(x){[y(x)\!+\!3x^2]}^2}
{{[1-x-3y(x)^2]}^3}
\end{align*}
Remark 1. We have now computed \(y''(x)\) — sort of. The answer is in terms of \(y(x)\text{,}\) which we don’t know. Since we cannot get an explicit formula for \(y(x)\text{,}\) there’s not a great deal that we can do, in general.
Remark 2. Even though we cannot solve
\(y=y^3+xy+x^3\) explicitly for
\(y(x)\text{,}\) for general
\(x\text{,}\) it is sometimes possible to solve equations like this for some special values of
\(x\text{.}\) In fact, we saw in Example
2.11.1 that when
\(x=1\text{,}\) the given equation reduces to
\(y(1)=y(1)^3+1\cdot y(1)+1^3\text{,}\) or
\(y(1)^3=-1\text{,}\) which we can solve to get
\(y(1)=-1\text{.}\) Substituting into (E2), as we did in Example
2.11.1 gives
\begin{equation*}
y'(1) = \frac{-1+3}{1-1-3(-1)^2} = -\frac{2}{3}
\end{equation*}
and substituting into (E4) gives
\begin{equation*}
y''(1) = \frac{6+2\big(-\frac{2}{3}\big)+6(-1)\big(-\frac{2}{3}\big)^2}
{1-1-3(-1)^2}
=\frac{6-\frac{4}{3}-\frac{8}{3}}{-3}
= -\frac{2}{3}
\end{equation*}
(It’s a fluke that, in this example, \(y'(1)\) and \(y''(1)\) happen to be equal.) So we now know that, even though we can’t solve \(y=y^3+xy+x^3\) explicitly for \(y(x)\text{,}\) the graph of the solution passes through \((1,-1)\) and has slope \(-\frac{2}{3}\) (i.e. is sloping downwards by between \(30^\circ\) and \(45^\circ\)) there and, furthermore, the slope of the graph decreases as \(x\) increases through \(x=1\text{.}\)
Here is a sketch of the part of the graph very near \((1, -1)\text{.}\) The tangent line to the graph at \((1, -1)\) is also shown. Note that the tangent line is sloping down to the right, as we expect, and that the graph lies below the tangent line near \((1,-1)\text{.}\) That’s because the slope \(f'(x)\) is decreasing (becoming more negative) as \(x\) passes through \(1\text{.}\)
