Integrate a chain-rule derivative first · source reference

Printed question

It is given that $\frac{dy}{dx}=\frac{e^{\tan^{-1}x}}{1+x^2}$, where $\tan^{-1}x$ denotes the principal value.

(i) Find an expression for $y$ in terms of $x$ given that $y=1$ when $x=0$. [2]

(ii) Show that $(1+x^2)\frac{d^2y}{dx^2}+(2x-1)\frac{dy}{dx}=0$. [2]

(iii) By further differentiation of the result in (ii), find the Maclaurin’s series for $y$ up to and including the term in $x^3$. [4]

(iv) Hence, deduce the series expansion for (a) $\frac{e^{\tan^{-1}x}}{(1+x)^2}$ (b) $e^{2x+\tan^{-1}x}$, up to and including the term in $x^2$. [4]

[2010/SAJC/Prelim P1/9]

Printed working

$\int y'\,dx=\int e^{\tan^{-1}x}/(1+x^2)\,dx$ gives $y=e^{\tan^{-1}x}+C$. At $x=0,y=1$: $1=e^0+C\Rightarrow C=0$.

$y'=y/(1+x^2)$; $(1+x^2)y'=y$; differentiate: $(1+x^2)y''+2xy'=y'\Rightarrow(1+x^2)y''+(2x-1)y'=0$.

Differentiate again: $(1+x^2)y'''+2xy''+(2x-1)y''+2y'=0$, so $(1+x^2)y'''+(4x-1)y''+2y'=0$.

At 0, $y=1$, $y'=1$, $y''=1$, $y'''=-1$; $y=1+x+x^2/2-x^3/6+\cdots$.

(a) $(1+x+x^2/2-x^3/6+\cdots)(1-2x+3x^2+\cdots)=1-x+3x^2/2+\cdots$.

(b) $e^{2x}e^{\tan^{-1}x}=[1+2x+(2x)^2/2!+\cdots][1+x+x^2/2-x^3/6+\cdots]=1+3x+9x^2/2+\cdots$.

Teacher note 1

Use binomial expansion for $(1+x)^{-2}$.

Teacher note 2

From MF27 use $e^x=1+x+x^2/2!+\cdots$, replacing $x$ by $2x$.

Mathematical review

(i) $y=e^{\tan^{-1}x}+C$; the initial value gives $C=0$.

(ii) Differentiate $(1+x^2)y'=y$ to obtain $(1+x^2)y''+(2x-1)y'=0$.

(iii) A further derivative gives $(1+x^2)y^{(3)}+(4x-1)y''+2y'=0$. At zero, $y=y'=y''=1$ and $y^{(3)}=-1$. Hence $y=1+x+x^2/2-x^3/6+O(x^4)$.

(iv)(a) Multiply by $(1+x)^{-2}=1-2x+3x^2+\cdots$ to obtain $1-x+3x^2/2+O(x^3)$. (b) Multiply by $e^{2x}=1+2x+2x^2+\cdots$ to obtain $1+3x+9x^2/2+O(x^3)$.

Provenance

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