Exams Joint Entrance Examination Mathematics Differential Equations Question 72 Question 72 of 490 Medium NUMERICAL
Let y = y ( x ) y=y(x) y = y ( x ) be the solution of the differential equation
2 cos x d y d x = sin 2 x − 4 y sin x , x ∈ ( 0 , π 2 ) 2\cos x\frac{dy}{dx}=\sin 2x-4y\sin x,x\in (0,\frac{\pi }{2}) 2 cos x d x d y = sin 2 x − 4 y sin x , x ∈ ( 0 , 2 π ) . If y ( π 3 ) = 0 y(\frac{\pi }{3})=0 y ( 3 π ) = 0 , then y ′ ( π 4 ) + y ( π 4 ) {y}^{'}(\frac{\pi }{4})+y(\frac{\pi }{4}) y ′ ( 4 π ) + y ( 4 π ) is equal to \_\_\_\_\_\_\_\_\_.
Answer: 1
Solutions · 1 Post your solution Official solution
d y d x + 2 y tan x = sin x = e 2 ∫ tan x d x = sec 2 x y sec 2 x = ∫ sin x cos 2 x d x = ∫ tan x sec x d x = sec x + C C = − 2 y = cos x − 2 cos 2 x y ( π 4 ) = 1 2 − 1 y ′ = − sin x + 4 cos 2 x sin x y ′ ( π 4 ) = − 1 2 + 2 y ′ ( π 4 ) + y ( π 4 ) = 1 \begin{gathered} \frac{dy}{dx}+2y\tan x=\sin x \\ ={e}^{2\int \tan xdx}={\sec }^{2}x \\ y{\sec }^{2}x=\int \frac{\sin x}{{\cos }^{2}x}dx \\ =\int \tan x\sec xdx \\ =\sec x+C \\ C=-2 \\ y=\cos x-2{\cos }^{2}x \\ y(\frac{\pi }{4})=\frac{1}{\sqrt{2}}-1 \\ {y}^{'}=-\sin x+4{\cos }^{2}x\sin x \\ {y}^{'}(\frac{\pi }{4})=-\frac{1}{\sqrt{2}}+2 \\ {y}^{'}(\frac{\pi }{4})+y(\frac{\pi }{4})=1 \end{gathered} d x d y + 2 y tan x = sin x = e 2 ∫ t a n x d x = sec 2 x y sec 2 x = ∫ cos 2 x sin x d x = ∫ tan x sec x d x = sec x + C C = − 2 y = cos x − 2 cos 2 x y ( 4 π ) = 2 1 − 1 y ′ = − sin x + 4 cos 2 x sin x y ′ ( 4 π ) = − 2 1 + 2 y ′ ( 4 π ) + y ( 4 π ) = 1 Answer the question to unlock