# Derive the particular solution if y'=ytanx-secx; y(0)=1.

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### 1 Answer

You should use the integrating factor method to solve o.d.e.

You need to remember the form of first order o.d.e. such that:

`y'+p(x)*y=q(x)`

Comapring the standard form to your equation yields `p(x)=-tan x ` and `q(x) = -sec x` .

You need to find the integrating factor, hence `I(x) = e^int -tan x dx`

You need to evaluate the indefinite integral `int - tan x dx` using substitution method such that:

`int -tan x dx = - int sin x/cos x dx`

You need to come up with the substitution `cos x= y =gt - sin x dx = dy`

`- int sin x/cos x dx = int (dy)/y = ln|y| + c = ln|cos x| + c`

Hence, hence `I(x) = e^(ln|cos x|) = cos x`

You need to multiply the equation by I(x) such that:

`y'*cos x - y*tan x*cos x = -sec x* cos x`

Reducing by cos x yields:

`y'*cos x - y*sin x = -1`

Notice that you may write the left side such that `(y*cos x)'.`

Integrating both sides yields:

`int (y*cos x)' dx = int -1 dx`

`` `y*cos x = -x + c`

You need to replace x by 0 such that:

`y(0)*cos 0 = 0 + c =gt 1*cos 0 = c =gt c = 1`

`y*cos x = -x + 1 =gt y = (1-x)/cos x`

**Hence, evaluating the particular solution to o.d.e. yields`y = (1-x)/cos x.` **