Question

solve this given differential equation​

Answer 1

Answer:

$\sf y=tan\:x-1+C\:e^{-tan\:x}$

Step-by-step explanation:

Given:

$\sf cos^2x\: \dfrac{dy}{dx} +y=tan\:x$

To Find:

The solution of the differential equation

Solution:

$\sf cos^2x\: \dfrac{dy}{dx} +y=tan\:x$

Dividing the whole equation by cos²x we get,

$\sf \dfrac{dy}{dx} +\dfrac{y}{cos^2x} =\dfrac{tan\:x}{cos^2x}$

$\sf \dfrac{dy}{dx} +y\:sec^2x=tan\:x\: sec^2\:x$

This is in the form of a linear differential equation,

$\sf \dfrac{dy}{dx} +Py=Q$

where P = sec²x and Q = tan x sec²x

Finding the integrating factor given by,

$\sf I.F=e^{\int\limits {P} \, dx }$

$\sf \implies e^{\int\limits {sec^2x} \, dx }$

$\sf \implies e^{tan\:x} (\because \int\limits{sec^2x} \, dx =tan\:x+C)$

Now the solution of the differential equation is given as,

$\sf y\times (I.F)=\displaystyle \sf \int\limits ({Q\times I.F)} \, dx +C$

Hence,

$\sf y\times e^{tan\:x}=\displaystyle \sf \int\limits {(tan\:x\:sec^2x\times e^{tan\:x})} \, dx$

Now let t = tan x

Differentiate on both sides w.r.t to x,

dt = sec²x dx

Substituting in the above equation,

$\sf y\times e^{tan\:x}=\displaystyle \sf \int\limits {(t\: e^{t})} \, dt$

Integrating by parts,

$\sf y\times e^{tan\:x}=t\displaystyle \sf \int\limits {e^t}\, dt -\bigg(1\times \int\limits {e^t}\bigg) \, dt$

$\sf y\times e^{tan\:x}=t\:e^t-e^t+C$

$\sf y\times e^{tan\:x}=e^t(t-1)+C$

$\sf y=\dfrac{e^t(t-1)}{e^{tan\:x}} +C\:e^{-tan\:x}$

Substitute the value of t,

$\sf y=\dfrac{e^{tan\:x}(tan\:x-1)}{e^{tan\:x}} +C\:e^{-tan\:x}$

$\sf y=tan\:x-1+C\:e^{-tan\:x}$