Exams Joint Entrance Examination Mathematics Differential Equations Question 70 Question 70 of 490 Medium NUMERICAL
Let y = y ( x ) y=y(x) y = y ( x ) be the solution of the differential equation d y d x + 2 y sec 2 x = 2 sec 2 x + 3 tan x ⋅ sec 2 x \frac{dy}{dx}+2y{\sec }^{2}x=2{\sec }^{2}x+3\tan x\cdot {\sec }^{2}x d x d y + 2 y sec 2 x = 2 sec 2 x + 3 tan x ⋅ sec 2 x such that y ( 0 ) = 5 4 y(0)=\frac{5}{4} y ( 0 ) = 4 5 . Then 12 ( y ( π 4 ) − e − 2 ) 12(y(\frac{\pi }{4})-{e}^{-2}) 12 ( y ( 4 π ) − e − 2 ) is equal to\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
Answer: 21
Solutions · 1 Post your solution Official solution
d y d x + 2 y sec 2 x = 2 sec 2 x + 3 tan x sec 2 x = e ∫ 2 sec 2 x d x \begin{gathered} \frac{dy}{dx}+2y{\sec }^{2}x=2{\sec }^{2}x+3\tan x{\sec }^{2}x \\ ={e}^{\int 2{\sec }^{2}xdx} \end{gathered} d x d y + 2 y sec 2 x = 2 sec 2 x + 3 tan x sec 2 x = e ∫ 2 s e c 2 x d x
= e 2 tan x y ⋅ e 2 tan x = ∫ e 2 tan x ( 2 + 3 tan x ) sec 2 x d x tan x = u sec 2 x d x = d u y ⋅ e 2 u = ∫ e 2 u ( 2 + 3 u ) d u y ⋅ e 2 u ⇒ 2 e 2 u 2 + 3 ∫ e 2 u ⋅ u d u y ⋅ e 2 u = e 2 u + 3 [ u e 2 u 2 − ∫ e 2 u 2 ] y e 2 u = e 2 u + 3 [ u e 2 u 2 − e 2 u 4 ] + C y e 2 tan x = e 2 tan x + 3 [ tan x e 2 tan x 2 − e 2 tan x 4 ] + C \begin{gathered} ={e}^{2\tan x} \\ y\cdot {e}^{2\tan x}=\int {e}^{2\tan x}(2+3\tan x){\sec }^{2}xdx \\ \tan x=u \\ {\sec }^{2}xdx=du \\ y\cdot {e}^{2u}=\int {e}^{2u}(2+3u)du \\ y\cdot {e}^{2u}\Rightarrow \frac{2{e}^{2u}}{2}+3\int {e}^{2u}\cdot udu \\ y\cdot {e}^{2u}={e}^{2u}+3[\frac{u{e}^{2u}}{2}-\int \frac{{e}^{2u}}{2}] \\ y{e}^{2u}={e}^{2u}+3[\frac{u{e}^{2u}}{2}-\frac{{e}^{2u}}{4}]+C \\ y{e}^{2\tan x}={e}^{2\tan x}+3[\frac{\tan x{e}^{2\tan x}}{2}-\frac{{e}^{2\tan x}}{4}]+C \end{gathered} = e 2 t a n x y ⋅ e 2 t a n x = ∫ e 2 t a n x ( 2 + 3 tan x ) sec 2 x d x tan x = u sec 2 x d x = d u y ⋅ e 2 u = ∫ e 2 u ( 2 + 3 u ) d u y ⋅ e 2 u ⇒ 2 2 e 2 u + 3 ∫ e 2 u ⋅ u d u y ⋅ e 2 u = e 2 u + 3 [ 2 u e 2 u − ∫ 2 e 2 u ] y e 2 u = e 2 u + 3 [ 2 u e 2 u − 4 e 2 u ] + C y e 2 t a n x = e 2 t a n x + 3 [ 2 tan x e 2 t a n x − 4 e 2 t a n x ] + C
F ( 0 ) = 5 4 5 4 = 1 − 3 4 + C 5 4 − 1 4 = C 1 = C y = 1 + 3 ( tan x 2 − 1 4 ) + 1 ⋅ e − 2 tan x y ( π 4 ) = 1 + 3 ( 1 2 − 1 4 ) + 1 e 2 y ( π 4 ) = 7 4 + 1 e 2 12 ( y ( x 4 ) − 1 e 2 ) = 12 ( 7 4 + 1 e 2 − 1 e 2 ) = 21 \begin{gathered} F(0)=\frac{5}{4} \\ \frac{5}{4}=1-\frac{3}{4}+C \\ \frac{5}{4}-\frac{1}{4}=C \\ 1=C \\ y=1+3(\frac{\tan x}{2}-\frac{1}{4})+1\cdot {e}^{-2\tan x} \\ y(\frac{\pi }{4})=1+3(\frac{1}{2}-\frac{1}{4})+\frac{1}{{e}^{2}} \\ y(\frac{\pi }{4})=\frac{7}{4}+\frac{1}{{e}^{2}} \\ 12(y(\frac{x}{4})-\frac{1}{{e}^{2}})=12(\frac{7}{4}+\frac{1}{{e}^{2}}-\frac{1}{{e}^{2}})=21 \end{gathered} F ( 0 ) = 4 5 4 5 = 1 − 4 3 + C 4 5 − 4 1 = C 1 = C y = 1 + 3 ( 2 tan x − 4 1 ) + 1 ⋅ e − 2 t a n x y ( 4 π ) = 1 + 3 ( 2 1 − 4 1 ) + e 2 1 y ( 4 π ) = 4 7 + e 2 1 12 ( y ( 4 x ) − e 2 1 ) = 12 ( 4 7 + e 2 1 − e 2 1 ) = 21 Answer the question to unlock