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