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{\bf Exam Question

Topic: DiffInt}

The function $f$ satisfies the equation
$$f(x)=\sqrt3+\int_0^x[1+(f(t))^2]\, dt.$$ By differentiating the
integral, find a differential equation for $f(x).$ Hence find the
function $f(x).$ \vspace{0.5in}

{\bf Solution}

$$f(x)]\sqrt3+\int_0^x[1+(f(t))^2]\, dt$$ Differentiating with
respect to $x$ gives $$\frac{df}{dx}=1+(f(x))^2\ \ \mathrm{so}\ \
\frac{1}{1+f^2}\frac{df}{dx}=1.$$

This gives $\tan^{-1}f=x+c,$ so $f(x)=\tan(x+c).$

Now $f(0)=\tan c=\sqrt3$ so $c=\pi/3$. Hence
$f(x)=\tan\left(x+\frac{\pi}{3}\right).$


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