Derivatives from a nonlinear differential identity · source reference

Printed question

The function $y=f(x)$ satisfies the differential equation $2y\frac{dy}{dx}=y^2-1$ and its graph passes the point $(0,3)$.

Show that when $x=0$, $\frac{d^2y}{dx^2}=\frac{20}{27}$. [3]

Find the Maclaurin’s series of $y$ in ascending powers of $x$ up to and including the term in $x^3$. [3]

[2008/RJC/Prelim /II/5(c)]

Printed working

$2yy'=y^2-1$ (1). Differentiate: $2yy''+2(y')^2=2yy'$; divide by 2: $yy''+(y')^2=yy'$ (2).

At 0, $y=3$, $y'=4/3$. Substituting in (2): $3y''+(4/3)^2=3(4/3)\Rightarrow y''=\frac13(4-16/9)=20/27$.

Differentiate (2): $yy'''+y'y''+2y'y''=(y')^2+yy''$; hence $yy'''+3y'y''=(y')^2+yy''$.

At 0, $y'''=28/81$. $y=3+(4/3)x+(20/27)x^2/2!+(28/81)x^3/3!+\cdots=3+4x/3+10x^2/27+14x^3/243+\cdots$.

Teacher note 1

Use implicit differentiation and product rule:

$$\frac d{dx}(yy')=y\frac d{dx}(y')+y'\frac d{dx}(y)=yy''+(y')^2.$$

Teacher note 2

Let $y=f(x)$. To find the Maclaurin’s series for $y$, up to and including the term in $x^3$, differentiate up to the 3rd derivative. Substitute the values of $y,y',y'',y'''$ when $x=0$ into Maclaurin’s theorem:

$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots.$$

Mathematical review

At zero, $6y'=8$, hence $y'(0)=4/3$. Differentiate: $2(y')^2+2yy''=2yy'$, or $(y')^2+yy''=yy'$. Thus $y''(0)=(4-16/9)/3=20/27$.

Differentiate again: $3y'y''+yy^{(3)}=(y')^2+yy''$. At zero, $3(4/3)(20/27)+3y^{(3)}=16/9+20/9=4$, so $y^{(3)}=28/81$. Therefore $y=3+4x/3+10x^2/27+14x^3/243+O(x^4)$. An independent check comes from $(y^2)'=y^2-1$, giving $y^2=1+8e^x$ and taking the positive branch near zero.

Provenance

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