If y (x) satisfies the differential equation y' – y tan x = 2x sec x and y(0) = 0, then

Engineering

Mathematics

Linear Differential Equation

Question

$y' (\frac{π}{3}) = \frac{π^{2}}{9}$

$y' (\frac{π}{4}) = \frac{π^{2}}{18}$

$y' (\frac{π}{3}) = \frac{4 π}{3} + \frac{2 π^{2}}{3 \sqrt{3}}$

$y (\frac{π}{4}) = \frac{π^{2}}{8 \sqrt{2}}$

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\frac{dy}{dx} - y tanx = 2 x (secx)$

I.F. = $e^{- \int tanx dx}$ = $e^{\ln \cos x}$ = cos x

$∴$ Solution is ycos x = $\int 2 xsecx cosx dx$

ycos x = x2 + C ; y(0) = 0 $\Rightarrow$ C = 0

$∴$ y = x² sec x and y' = 2x sec x + x² sec x tan x

$y (\frac{π}{4}) = \frac{π^{2}}{8 \sqrt{2}}; y' (\frac{π}{4}) = \frac{π}{\sqrt{2}} + \frac{π^{2}}{8 \sqrt{2}}$

$y (\frac{π}{3}) = \frac{2 π^{2}}{9}; y' (\frac{π}{3}) = \frac{4 π}{3} + \frac{2 π^{2}}{3 \sqrt{3}}$

Please subscribe our Youtube channel to unlock this solution.