Question

D^2+5D+6]y=e^(-2x)sec^2x(1+2tanx)
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Answer 1

Answer:

(D^2+5D+6) y=e^2xsec^x(1+2tanx)

Answer 2

Given:

$D^{2} +5 D+6 y=e^{-2 x} \sec ^{2} x(1+2 \tan x)$

To find:

Solution of the differential equation.

Solution:

$D^{2}+5 D+6=0 \\D=-3,-2 \\\text { C.F.is } y_{c}=c_{1} e^{-3 x}+c_{2} e^{-2 x} \\\text { P.I }=y_{p}=\frac{1}{D^{2}+5 D+6} e^{-2 x} \sec ^{2} x(1+2 \tan x) \\=\frac{1}{(D+2)(D+3)} e^{-2 x} \sec ^{2} x(1+2 \tan x) \\=\frac{1}{(D+3)}\left[e^{-2 x} \int e^{2 x} e^{-2 x} \sec ^{2} x(1+2 \tan x) d x\right] \\=\frac{1}{(D+3)}\left[e^{-2 x} \int \sec ^{2} x(1+2 \tan x) d x\right] \\\text { Put } \tan x=t \\\ \sec ^{2} x d x=d t$

$=\frac{1}{(D+3)}\left[e^{-2 x} \int(1+2 t) d t\right] \\=\frac{1}{(D+3)}\left[e^{-2 x}\left(t+t^{2}\right)\right] \\=\frac{1}{(D+3)}\left[e^{-2 x}\left(\tan x+\tan ^{2} x\right)\right] \\=e^{-3 x} \int\left[e^{3 x} e^{-2 x}\left(\tan x+\tan ^{2} x\right) d x\right] \\=e^{-3 x} \int\left[e^{x}\left(\tan x+\sec ^{2} x-1\right) d x\right]\end{array}$

$\begin{array}{l}=e^{-3 x}\left[\int\left[e^{x}\left(\tan x+\sec ^{2} x\right) d x-\int e^{x} d x\right]\right. \\=e^{-3 x}\left[e^{x} \tan x-e^{x}\right] \\=e^{-2 x}[\tan x-1]\end{array} \therefore The complete solution is \mathrm{y}=\mathrm{C} . \mathrm{F} . + P.I. \therefore y=c_{1} e^{-3 x}+c_{2} e^{-2 x}+e^{-2 x}[\tan x-1]$