If y(x) satisfy the differential equation dydx+yx=cos x+sin xx and y(0)=0 then

dydx+(1x)y=cosx+sinxxI.F=e∫1x dx=elogx=xSolution isy(I.F)=∫(cosx+sinxx)(I.F) dx⇒ xy= ∫(cosx+sinxx)x dx= ∫(x cosx+sinx) dx= xsinx− ∫1. sinx dx− cosx+c= x sinx+cosx−cosx+c⇒ y.x= x sinx+cPut x=0y(0)(0)= 0 sinx+c⇒ c=0∴ y.x= x sinxy= sinxy(π2)= sin(π2)=1y(π4)= sin(π4)= 12

If y(x) satisfy the differential equation dydx+yx=cos x+sin xx and y(0)=0 then

dydx+(1x)y=cosx+sinxxI.F=e∫1x dx=elogx=xSolution isy(I.F)=∫(cosx+sinxx)(I.F) dx⇒ xy= ∫(cosx+sinxx)x dx= ∫(x cosx+sinx) dx= xsinx− ∫1. sinx dx− cosx+c= x sinx+cosx−cosx+c⇒ y.x= x sinx+cPut x=0y(0)(0)= 0 sinx+c⇒ c=0∴ y.x= x sinxy= sinxy(π2)= sin(π2)=1y(π4)= sin(π4)= 12

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If y(x) satisfy the differential equation $\frac{dy}{dx} + \frac{y}{x} = \cos x + \frac{\sin x}{x}$ and y(0)=0 then

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$y (\frac{π}{4}) = \frac{1}{2 \sqrt{2}}$

$y (\frac{π}{2}) = \frac{1}{2}$

$y (\frac{π}{2}) = \frac{1}{4}$

$y (\frac{π}{4}) = \frac{1}{\sqrt{2}}$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} + (\frac{1}{x}) y = \cos x + \frac{\sin x}{x} \\ I . F = e^{\int \frac{1}{x} d x} = e^{\log x} = x \\ S o l u t i o n i s \\ y (I . F) = \int (\cos x + \frac{\sin x}{x}) (I . F) d x \\ \Rightarrow x y = \int (\cos x + \frac{\sin x}{x}) x d x \\ = \int (x \cos x + \sin x) d x \\ = x \sin x - \int 1. \sin x d x - \cos x + c \\ = x \sin x + \cos x - \cos x + c \\ \Rightarrow y . x = x \sin x + c \end{array}$

$\begin{array}{l} P u t x = 0 \\ y (0) (0) = 0 \sin x + c \\ \Rightarrow c = 0 \\ ∴ y . x = x \sin x \\ y = \sin x \\ y (\frac{π}{2}) = \sin (\frac{π}{2}) = 1 \\ y (\frac{π}{4}) = \sin (\frac{π}{4}) = \frac{1}{\sqrt{2}} \end{array}$