Let dydx - 2y x-cos x such that y -2 - 0- If the maximum value of y is k- then the value of k is

Dydx 2y x=cos x This is a linear differential equation in x. I.F. = e^ 2∫ x dx = 1sin^2x General solution is y sin^2x = ∫cos xsin^2x dx +C ⇒ y= sin x+Csin^ ...

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Standard XII

Mathematics

Relation between Continuity and Differentiability

Let dydx - 2y...

Question

Let $\frac{d y}{d x} - 2 y cot x = cos x$ such that $y (\frac{π}{2}) = 0$ . If the maximum value of $y$ is $k,$ then the value of $k$ is

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Solution

$\frac{d y}{d x} - 2 y cot x = cos x$
This is a linear differential equation in $x .$
$I.F. = e^{- 2 \int cot x d x} = \frac{1}{{sin}^{2} x}$
General solution is
$\frac{y}{{sin}^{2} x} = \int \frac{cos x}{{sin}^{2} x} d x + C$
$\Rightarrow y = - sin x + C {sin}^{2} x$
Since $y (\frac{π}{2}) = 0,$ therefore $C = 1$
$∴ y = {sin}^{2} x - sin x$
$\Rightarrow y = {(sin x - \frac{1}{2})}^{2} - \frac{1}{4}$
$∴$ Maximum value of $y$ is ${(- 1 - \frac{1}{2})}^{2} - \frac{1}{4} = 2$

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Let dydx−2ycotx=cosx such that y(π2)=0. If the maximum value of y is k, then the value of k is

dydx−2ycotx=cosx This is a linear differential equation in x. I.F.=e−2∫cotx dx=1sin2x General solution is ysin2x=∫cosxsin2xdx+C ⇒y=−sinx+Csin2x Since y(π2)=0, therefore C=1 ∴y=sin2x−sinx ⇒y=(sinx−12)2−14 ∴ Maximum value of y is (−1−12)2−14=2

Let $\frac{d y}{d x} - 2 y cot x = cos x$ such that $y (\frac{π}{2}) = 0$ . If the maximum value of $y$ is $k,$ then the value of $k$ is