Let y=y (x) be the solution of the differential equation dydx+2y2cos4 x− cos 2x= xetan−1(2 cot 2x),0<x<π2 with y(π4)=π232, If y(π3)=π218e−tan−1(α),then the value of 3α2 is equal to

dydx+22cos4x−cos2xy=xetan−1(2cot 2x)Here given differential equation is in the form of dydx+py=Q∫dx2cos4x−cos2x{∵P=∫dx2cos4x−cos2x=∫dxcos4x+sin4x=∫cosec4xdx1+cot4x=−∫t2+1t4+1dt= −∫(1+1t2)(t−1t)2+2dt= −12tan−1(t−1t2)cot x= t= −12tan−1(2cot 2x) ∴IF =e−tan−1(2cot 2x)ye−tan−1(2cot 2x)=∫xdx ; ye−tan−1(2cot 2x)=x22+cy(π4)=π232+c⇒c=0 ; y=x22etan−1(2cot 2x)y(π3)=π218etan−1 (2 cot 2π3)=π218e−tan−1 (23)α=23⇒3α2=2

Let y=y (x) be the solution of the differential equation dydx+2y2cos4 x− cos 2x= xetan−1(2 cot 2x),0<x<π2 with y(π4)=π232, If y(π3)=π218e−tan−1(α),then the value of 3α2 is equal to

dydx+22cos4x−cos2xy=xetan−1(2cot 2x)Here given differential equation is in the form of dydx+py=Q∫dx2cos4x−cos2x{∵P=∫dx2cos4x−cos2x=∫dxcos4x+sin4x=∫cosec4xdx1+cot4x=−∫t2+1t4+1dt= −∫(1+1t2)(t−1t)2+2dt= −12tan−1(t−1t2)cot x= t= −12tan−1(2cot 2x) ∴IF =e−tan−1(2cot 2x)ye−tan−1(2cot 2x)=∫xdx ; ye−tan−1(2cot 2x)=x22+cy(π4)=π232+c⇒c=0 ; y=x22etan−1(2cot 2x)y(π3)=π218etan−1 (2 cot 2π3)=π218e−tan−1 (23)α=23⇒3α2=2

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Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{\sqrt{2} y}{2 \cos^{4} x - \cos 2 x} = x e^{\tan^{- 1} (\sqrt{2} \cot 2 x)}, 0 < x < \frac{π}{2}$ with $y (\frac{π}{4}) = \frac{π^{2}}{32},$ If $y (\frac{π}{3}) = \frac{π^{2}}{18} e^{- \tan^{- 1}} (α),$ then the value of $3 α^{2}$ is equal to

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Detailed Solution

$\frac{d y}{d x} + \frac{\sqrt{2}}{2 \cos^{4} x - \cos 2 x} y = x e^{\tan - 1 (\sqrt{2} \cot 2 x)}$

Here given differential equation is in the form of $\frac{d y}{d x} + p y = Q$

$\begin{array}{l} \int \frac{d x}{2 \cos^{4} x - \cos 2 x} {∵ P = \int \frac{d x}{2 \cos^{4} x - \cos 2 x} \\ = \int \frac{d x}{\cos^{4} x + \sin^{4} x} = \int \frac{\cos e c^{4} x d x}{1 + \cot^{4} x} \end{array}$

$\begin{array}{l} = - \int \frac{t^{2} + 1}{t^{4} + 1} d t = - \int \frac{(1 + \frac{1}{t^{2}})}{{(t - \frac{1}{t})}^{2} + 2} d t = \frac{- 1}{\sqrt{2}} \tan^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) \\ \cot x = t \\ = - \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{2} \cot 2 x) ∴ I F = e^{- \tan^{- 1} (\sqrt{2} \cot 2 x)} \\ y e^{- \tan^{- 1} (\sqrt{2} \cot 2 x)} = \int x d x; y e^{- \tan^{- 1} (\sqrt{2} \cot 2 x)} = \frac{x^{2}}{2} + c \\ y (\frac{π}{4}) = \frac{π^{2}}{32} + c \Rightarrow c = 0; y = \frac{x^{2}}{2} e^{\tan^{- 1} (\sqrt{2} \cot 2 x)} \\ y (\frac{π}{3}) = \frac{π^{2}}{18} e^{\tan^{- 1} (\sqrt{2} \cot \frac{2 π}{3})} = \frac{π^{2}}{18} e^{- \tan^{- 1} (\sqrt{\frac{2}{3}})} \\ α = \sqrt{\frac{2}{3}} \Rightarrow 3 α^{2} = 2 \end{array}$