Exams Joint Entrance Examination Mathematics Differential Equations Question 82 Question 82 of 490 Medium NUMERICAL
Let y = y ( x ) y=y(x) y = y ( x ) be the solution of the differential equation
sec 2 x d x + ( e 2 y tan 2 x + tan x ) d y = 0 , 0 < x < π 2 , y ( π / 4 ) = 0 {\sec }^{2}xdx+({e}^{2y}{\tan }^{2}x+\tan x)dy=0,0<x<\frac{\pi }{2},y(\pi /4)=0 sec 2 x d x + ( e 2 y tan 2 x + tan x ) d y = 0 , 0 < x < 2 π , y ( π /4 ) = 0 .
If y ( π / 6 ) = α y(\pi /6)=\alpha y ( π /6 ) = α , then e 8 α {e}^{8\alpha } e 8 α is equal to \_\_\_\_\_\_\_\_\_\_\_\_.
Answer: 9
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sec 2 x d x d y + e 2 y tan 2 x + tan x = 0 ( tan x = t ⇒ sec 2 x d x d y = d t d y ) d t d y + e 2 y × t 2 + t = 0 d t d y + t = − t 2 ⋅ e 2 y 1 t 2 d t d y + 1 t = − e 2 y ( 1 t = u − 1 t 2 d t d y = d u d y ) − d u d y + u = − e 2 y d u d y − u = e 2 y = e − ∫ d y = e − y u e − y = ∫ e − y × e 2 y d y 1 tan x × e − y = e y + c x = π 4 , y = 0 , c = 0 x = π 6 , y = α 3 e − α = e α + 0 e 2 α = 3 e 8 α = 9 \begin{gathered} {\sec }^{2}x\frac{dx}{dy}+{e}^{2y}{\tan }^{2}x+\tan x=0 \\ (\tan x=t\Rightarrow {\sec }^{2}x\frac{dx}{dy}=\frac{dt}{dy}) \\ \frac{dt}{dy}+{e}^{2y}\times {t}^{2}+t=0 \\ \frac{dt}{dy}+t=-{t}^{2}\cdot {e}^{2y} \\ \frac{1}{{t}^{2}}\frac{dt}{dy}+\frac{1}{t}=-{e}^{2y} \\ (\frac{1}{t}=u\frac{-1}{{t}^{2}}\frac{dt}{dy}=\frac{du}{dy}) \\ \frac{-du}{dy}+u=-{e}^{2y} \\ \frac{du}{dy}-u={e}^{2y} \\ ={e}^{-\int dy}={e}^{-y} \\ u{e}^{-y}=\int {e}^{-y}\times {e}^{2y}dy \\ \frac{1}{\tan x}\times {e}^{-y}={e}^{y}+c \\ x=\frac{\pi }{4},y=0,c=0 \\ x=\frac{\pi }{6},\, y=\alpha \\ \sqrt{3}{e}^{-\alpha }={e}^{\alpha }+0 \\ {e}^{2\alpha }=\sqrt{3} \\ {e}^{8\alpha }=9 \end{gathered} sec 2 x d y d x + e 2 y tan 2 x + tan x = 0 ( tan x = t ⇒ sec 2 x d y d x = d y d t ) d y d t + e 2 y × t 2 + t = 0 d y d t + t = − t 2 ⋅ e 2 y t 2 1 d y d t + t 1 = − e 2 y ( t 1 = u t 2 − 1 d y d t = d y d u ) d y − d u + u = − e 2 y d y d u − u = e 2 y = e − ∫ d y = e − y u e − y = ∫ e − y × e 2 y d y tan x 1 × e − y = e y + c x = 4 π , y = 0 , c = 0 x = 6 π , y = α 3 e − α = e α + 0 e 2 α = 3 e 8 α = 9 Answer the question to unlock