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Question Number 140896 by bramlexs22 last updated on 14/May/21

the function f with variable x

satisfies the equation

x^{2} f^{'} (x) + 2 x f (x) = \arctan x for

0 < \arctan x < \frac{π}{2} and f (1) = \frac{π}{4} .

find f (x) .

Answered by EDWIN88 last updated on 15/May/21

let g (x) = x^{2} f (x), diff wrt x give

g^{'} (x) = x^{2} f^{'} (x) + 2 x f (x)

so that g^{'} (x) = \arctan x

integration by part

g (x) = \int \arctan x dx = x \arctan x - \frac{1}{2} \ln (1 + x^{2}) + c

the condition g (1) = f (1) = \frac{π}{4}

\Rightarrow \frac{π}{4} = \arctan 1 - \frac{1}{2} \ln (2) + c

\Rightarrow c = \ln \sqrt{2}

∴ f (x) = \frac{g (x)}{x^{2}} = \frac{\arctan x}{x} - \frac{\ln \sqrt{1 + x^{2}}}{x^{2}} + \frac{\ln \sqrt{2}}{x^{2}}

Answered by mathmax by abdo last updated on 14/May/21

f (x) = y e \Rightarrow x^{2} y^{^{'}} + 2 xy = arctanx and y (1) = \frac{π}{4}

h \to {xy}^{^{'}} + 2 y = 0 \Rightarrow {xy}^{^{'}} = - 2 y \Rightarrow \frac{y^{^{'}}}{y} = - \frac{2}{x} \Rightarrow \ln ∣ y ∣= - 2 \log ∣ x ∣ + c \Rightarrow

y = \frac{k}{x^{2}} mvc method \to y^{^{'}} = \frac{k^{^{'}}}{x^{2}} + k (\frac{- 2 x}{x^{4}}) = \frac{k^{^{'}}}{x^{2}} - \frac{2 k}{x^{3}}

e \Rightarrow k^{^{'}} - \frac{2 k}{x} + \frac{2 k}{x} = arctanx \Rightarrow k = \int arctanx dx

= xarctanx - \int \frac{x}{1 + x^{2}} dx = xarctan (x) - \frac{1}{2} \log (1 + x^{2}) + C \Rightarrow

y (x) = \frac{1}{x^{2}} (xarctan (x) - \frac{1}{2} \log (1 + x^{2}) + C)

= \frac{\arctan (x)}{x} - \frac{1}{{2 x}^{2}} \log (1 + x^{2}) + \frac{c}{x^{2}} = f (x)

f (1) = \frac{π}{4} \Rightarrow \frac{π}{4} - \frac{1}{2} \log 2 + C = \frac{π}{4} \Rightarrow C = \frac{1}{2} \log 2 \Rightarrow

f (x) = \frac{\arctan (x)}{x} - \frac{1}{{2 x}^{2}} \log (1 + x^{2}) + \frac{\log 2}{{2 x}^{2}}