A function $$y=f(x)$$ satisfies the differential equation $$f(x)$$·$$sin2x-cosx+1+sin2xf$$'$$(x)=0$$ with initial condition $$y(0)=0$$ . The value of $$f($$π$$/6)=$$

Methods of solving first order first degree differential equations

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$A function y = f (x) satisfies the differential equation f (x) \cdot \sin 2 x - \cos x + (1 + \sin^{2} x) f^{'} (x) = 0$ $with initial condition y (0) = 0 . The value of f (π / 6) =$

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Solution

$y \sin 2 x - \cos x + (1 + \sin^{2} x) \frac{d y}{d x} = 0$
$\frac{d y}{d x} + \frac{\sin 2 x}{1 + \sin^{2} x} = \frac{\cos x}{1 + \sin^{2} x}$
$I . F = e^{\int \frac{\sin 2 x}{1 + \sin^{2} x} d x} = e^{\log (1 + \sin^{2} x)} = 1 + \sin^{2} x$
Solution is
$y (1 + \sin^{2} x) = \int \frac{\cos x}{(1 + \sin^{2} x)} (1 + \sin^{2} x) d x$
$y (1 + \sin^{2} x) = \sin x + C [y (0) = 0 \Rightarrow C = 0]$
$y = \frac{\sin x}{1 + \sin^{2} x} ∴ y (π / 6) = \frac{2}{5} = 0.4$

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Methods of solving first order first degree differential equations

Remember concepts with our Masterclasses.

A function y=f(x) satisfies the differential equation f(x)·sin2x-cosx+1+sin2xf'(x)=0 with initial condition y(0)=0 . The value of f(π/6)=

ysin2x−cosx+1+sin2xdydx=0dydx+sin2x1+sin2x=cosx1+sin2xI.F=e∫sin2x1+sin2xdx=elog1+sin2x=1+sin2xSolution isy1+sin2x=∫cosx1+sin2x1+sin2xdxy1+sin2x=sinx+C[y(0)=0⇒C=0]y=sinx1+sin2x ∴yπ/6=25=0.4

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$A function y = f (x) satisfies the differential equation f (x) \cdot \sin 2 x - \cos x + (1 + \sin^{2} x) f^{'} (x) = 0$ $with initial condition y (0) = 0 . The value of f (π / 6) =$