If a function $$y=f(x)$$ satisfies the differential equation $$f(x)$$⋅$$sin$$⁡$$2x$$−$$cos$$⁡$$x+1+sin2$$⁡xf′$$(x)=0$$ with initial condition $$y(0)=0$$, then the value of f$$π$$$$6$$ is equal to

Question

If a function $$y=f(x)$$ satisfies the differential equation $$f(x)$$⋅$$sin$$⁡$$2x$$−$$cos$$⁡$$x+1+sin2$$⁡xf′$$(x)=0$$ with initial condition $$y(0)=0$$, then the value  of f$$π$$$$6$$ is equal to

Infinity Learn · Accepted Answer

The given equation is $$f(x)$$⋅$$sin$$⁡$$2x$$−$$cos$$⁡$$x+1+sin2$$⁡xf′(x)=0$$⇒ysin⁡$$2x$$−$$cos$$⁡$$x+1+sin2$$⁡$$xdydx=0$$⇒$$dydx+$$$$sin$$⁡$$2x1+sin2$$⁡$$xy=$$$$cos$$⁡$$x1+sin2$$⁡x is in the form of $$dydx+Py=Q$$ Where $$P=$$$$sin$$⁡$$2x1+sin2$$⁡x and $$Q=$$$$cos$$⁡$$x1+sin2$$⁡x Integrating Factor $$(IF)=e$$∫$$sin$$⁡$$2x1+sin2$$⁡$$xdx=eln$$⁡$$1+sin2$$⁡x ∵∫f′$$(x)f(x)dx=ln$$⁡|$$f(x)|$$+C=1+sin2$$⁡x General solution is y⋅(IF)=$$∫$$(IF)Qdx$$⇒$$y1+sin2$$⁡$$x=$$∫$$1+sin2$$⁡xcos⁡$$x1+sin2$$⁡xdx⇒$$y1+sin2$$⁡$$x=$$∫$$cos$$⁡$$xdx=$$$$sin$$⁡$$x+C$$ When $$x=0$$,$$y=0$$⇒$$C=0$$ When $$x=$$π$$6$$, then $$y1+14=12$$⇒$$y54=12$$⇒$$y=25$$∴f$$π$$$$6=25Therefore$$, the correct answer is $$(4).

$If a function y = f (x) satisfies the differential equation$ $f (x) \cdot \sin 2 x - \cos x + (1 + \sin^{2} x) f^{'} (x) = 0 with initial condition y (0) = 0, then the value$ $of f (\frac{π}{6}) is equal to$

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If a function y=f(x) satisfies the differential equation f(x)⋅sin⁡2x−cos⁡x+1+sin2⁡xf′(x)=0 with initial condition y(0)=0, then the value of fπ6 is equal to

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$If a function y = f (x) satisfies the differential equation$ $f (x) \cdot \sin 2 x - \cos x + (1 + \sin^{2} x) f^{'} (x) = 0 with initial condition y (0) = 0, then the value$ $of f (\frac{π}{6}) is equal to$