Exams Joint Entrance Examination Mathematics Differential Equations Question 488 Question 488 of 490 Medium MCQ
Let y = y ( x ) y=y(x) y = y ( x ) be the solution of the differential equation
d y d x + 3 ( tan 2 x ) y + 3 y = sec 2 x , y ( 0 ) = 1 3 + e 3 \frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x, y(0)=\frac{1}{3}+e^3 d x d y + 3 ( tan 2 x ) y + 3 y = sec 2 x , y ( 0 ) = 3 1 + e 3 . Then y ( π 4 ) y\left(\frac{\pi}{4}\right) y ( 4 π ) is equal to :
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\[ d y d x + 3 ( tan 2 x ) y + 3 y = sec 2 x ⇒ d y d x + 3 sec 2 x y = sec 2 x I.F = e ∫ 3 sec 2 x d x = e 3 tan x \begin{aligned}
& \frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x \\
& \Rightarrow \frac{d y}{d x}+3 \sec ^2 x y=\sec ^2 x \\
& \text { I.F }=e^{\int 3 \sec ^2 x d x} \\
& \quad=e^{3 \tan x}
\end{aligned} d x d y + 3 ( tan 2 x ) y + 3 y = sec 2 x ⇒ d x d y + 3 sec 2 x y = sec 2 x I.F = e ∫ 3 s e c 2 x d x = e 3 t a n x \]
\[ y ⋅ e tan x = ∫ e 3 tan x ⋅ sec 2 x d x + c y ⋅ e 3 tan x = e 3 tan x 3 + c Also f ( 0 ) = 1 3 + e 3 ⇒ ( 1 3 + e 3 ) = 1 3 + c ⇒ c = e 3 ∴ y ⋅ e 3 tan x = e 3 tan x 3 + e 3 Put x = π 4 \begin{aligned}
& y \cdot e^{\tan x}=\int e^{3 \tan x} \cdot \sec ^2 x d x+c \\
& y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+c \\
& \text { Also } f(0)=\frac{1}{3}+e^3 \\
& \Rightarrow\left(\frac{1}{3}+e^3\right)=\frac{1}{3}+c \\
& \Rightarrow c=e^3 \\
& \therefore y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+e^3 \\
& \text { Put } x=\frac{\pi}{4}
\end{aligned} y ⋅ e t a n x = ∫ e 3 t a n x ⋅ sec 2 x d x + c y ⋅ e 3 t a n x = 3 e 3 t a n x + c Also f ( 0 ) = 3 1 + e 3 ⇒ ( 3 1 + e 3 ) = 3 1 + c ⇒ c = e 3 ∴ y ⋅ e 3 t a n x = 3 e 3 t a n x + e 3 Put x = 4 π \]
y e 3 = e 3 3 + e 3 ⇒ y = 4 3 y e^3=\frac{e^3}{3}+e^3 \Rightarrow y=\frac{4}{3} y e 3 = 3 e 3 + e 3 ⇒ y = 3 4 Answer the question to unlock