Let y = y(x) be the solution of the differential equation, + y tan x = 2x + x2 tan x, , such that y(0) = 1. Then

Engineering

Mathematics

Linear Differential Equation

Question

Let y = y(x) be the solution of the differential equation, $\frac{dy}{dx}$ + y tan x = 2x + x² tan x, $x \in (- \frac{π}{2}, \frac{π}{2})$ , such that y(0) = 1. Then

$y' (\frac{π}{4}) + y' (\frac{- π}{4}) = - \sqrt{2}$

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

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

$y (\frac{π}{4}) + y (- \frac{π}{4}) = \frac{π^{2}}{2} + 2$

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$\frac{dy}{dx} + y (tanx) = 2 x + x^{2} tanx$

$I . F . = e^{\pm \int tanx dx} = e^{\ln . secx} = secx$

$∴$ y.sec x = $\int$ (2x + x² tan x)sec x.dx

= $\int$ 2x sec x dx + $\int$ x² (sec x.tan x)dx

y sec x = x² sec x + λ

⇒ y = x² + λ cos x

y (0) = 0 + λ = 1 ⇒ λ = 1

y = x² + cos x

$y (\frac{π}{4}) = \frac{π^{2}}{16} + \frac{1}{\sqrt{2}}$

$y (- \frac{π}{4}) = \frac{π^{2}}{16} + \frac{1}{\sqrt{2}}$

y '(x) = 2x – sin x

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

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

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