Solve-the-differential-equation-cos-2-x-dy-dx-y-tan-x-

Question Number 5991 by sanusihammed last updated on 08/Jun/16

$${Solve}\:{the}\:{differential}\:{equation}\: \\ $$$$ \\ $$$${cos}^{\mathrm{2}} \left({x}\right)\:\frac{{dy}}{{dx}}\:+\:{y}\:=\:{tan}\left({x}\right) \\ $$

Answered by prakash jain last updated on 09/Jun/16

(dy/dx)+ysec^2 x=tan xsec^2 x equation of the form y′+P(x)y=Q(x) integrating factor u(x)=e^(∫P(x)dx) =e^(tan x) y=((∫e^(tan x) tan xsec^2 x dx +C)/e^(tan x) ) integral ∫e^(tan x) tan xsec^2 x dx can be evaluated by substituting tan x=u

$$\frac{{dy}}{{dx}}+{y}\mathrm{sec}^{\mathrm{2}} {x}=\mathrm{tan}\:{x}\mathrm{sec}^{\mathrm{2}} {x} \\ $$$${equation}\:{of}\:{the}\:{form} \\ $$$${y}'+{P}\left({x}\right){y}={Q}\left({x}\right) \\ $$$${integrating}\:{factor}\:{u}\left({x}\right)={e}^{\int{P}\left({x}\right){dx}} \\ $$$$={e}^{\mathrm{tan}\:{x}} \\ $$$${y}=\frac{\int{e}^{\mathrm{tan}\:{x}} \:\mathrm{tan}\:{x}\mathrm{sec}^{\mathrm{2}} {x}\:{dx}\:+{C}}{{e}^{\mathrm{tan}\:{x}} } \\ $$$${integral}\:\int{e}^{\mathrm{tan}\:{x}} \mathrm{tan}\:{x}\mathrm{sec}^{\mathrm{2}} {x}\:{dx} \\ $$$${can}\:{be}\:{evaluated}\:{by}\:{substituting}\:\mathrm{tan}\:{x}={u} \\ $$

Commented by prakash jain last updated on 09/Jun/16

A differential equation of the form y′+P(x)y=Q(x) has the following solution y=((∫Q(x)e^(∫P(x)dx) dx+C)/e^(∫P(x)dx) ) e^(∫P(x)dx) is called integrating factor.

$$\mathrm{A}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form} \\ $$$${y}'+{P}\left({x}\right){y}={Q}\left({x}\right) \\ $$$$\mathrm{has}\:\mathrm{the}\:\mathrm{following}\:\mathrm{solution} \\ $$$${y}=\frac{\int{Q}\left({x}\right){e}^{\int{P}\left({x}\right){dx}} {dx}+{C}}{{e}^{\int{P}\left({x}\right){dx}} } \\ $$$${e}^{\int{P}\left({x}\right){dx}} \:\mathrm{is}\:\mathrm{called}\:\mathrm{integrating}\:\mathrm{factor}. \\ $$

Commented by prakash jain last updated on 09/Jun/16

∫e^(tan x) tan xsec^2 x =∫ue^u du=e^u (u−1) =e^(tan x) (tan x−1) Solution y=((e^(tan x) (tan x−1)+C)/e^(tan x) ) =tan x−1+Ce^(tan x)

$$\int{e}^{\mathrm{tan}\:{x}} \mathrm{tan}\:{x}\mathrm{sec}^{\mathrm{2}} {x}\:=\int{ue}^{{u}} {du}={e}^{{u}} \left({u}−\mathrm{1}\right) \\ $$$$={e}^{\mathrm{tan}\:{x}} \left(\mathrm{tan}\:{x}−\mathrm{1}\right) \\ $$$$\mathrm{Solution} \\ $$$${y}=\frac{{e}^{\mathrm{tan}\:{x}} \left(\mathrm{tan}\:{x}−\mathrm{1}\right)+{C}}{{e}^{\mathrm{tan}\:{x}} } \\ $$$$=\mathrm{tan}\:{x}−\mathrm{1}+{Ce}^{\mathrm{tan}\:{x}} \\ $$

Commented by sanusihammed last updated on 08/Jun/16

$${Please}\:{i}\:{dont}\:{get}\:{you} \\ $$$$ \\ $$

Commented by sanusihammed last updated on 09/Jun/16

$${kindly}\:{show}\:{the}\:{steps}\:.\:{thanks} \\ $$

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