Let y = y(x) be the solution of the differential equation dy/dx+3(tan^2 x)y+3y=sec^2x, y(0)=1/3+e^3.

Question

Let y = y(x) be the solution of the differential equation \(\frac{d y}{d x}+3\left(\tan ^{2} x\right) y+3 y=\sec ^{2} x, y(0)=\frac{1}{3}+e^{3}.\) Then \(y\left(\frac{\pi}{4}\right)\) is equal to

(1) \(\frac{2}{3}\)

(2) \(\frac{4}{3}\)

(3) \(\frac{4}{3}+\mathrm{e}^{3}\)

(4) \(\frac{2}{3}+\mathrm{e}^{3}\)

Answer 1

Correct option is (2) \(\frac{4}{3}\) 

\(\frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x\)

\( \Rightarrow \frac{d y}{d x}+3 \sec ^2 x y=\sec ^2 x \)

\(\text { I.F }=e^{\int 3 \sec ^2 x d x} \)

\(=e^{3 \tan x} \)

\( y \cdot e^{\tan x}=\int e^{3 \tan x} \cdot \sec ^2 x d x+c \)

\( y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+c\)

Also \(f(0)=\frac{1}{3}+e^3\)

\( \Rightarrow\left(\frac{1}{3}+e^3\right)=\frac{1}{3}+c \)

\( \Rightarrow c=e^3 \)

\( \therefore y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+e^3\)

Put \(x=\frac{\pi}{4}\)

\(y e^3=\frac{e^3}{3}+e^3 \Rightarrow y=\frac{4}{3}\)