Question

Solve the following initial-value problem: cos(x) dy/dx + sin(x) y - 1 = 0, y(0) = 1

Answer 1

y= x + cosx

Explanation:

`cosx\frac{\mathrm{d}y }{\mathrm{d} x}+(sinx)y-1=0\\cosx\frac{\mathrm{d}y }{\mathrm{d} x}+(sinx)y=1\\\frac{\mathrm{d}y }{\mathrm{d} x}+(tanx)y=secx`

now equation is in linear differential form

finding integrating factor;

I.F. =
`e^(\int tanx dx) = e^(ln\ secx) = secx`

`y=(1)/(I.F)(\int Q dx + c)`

`y=(1)/(secx)(\int secx dx + c)`

`y=\int  dx + c(cosx)\\y= x + c(cosx)`

using y(0) = 1

1 = 0 + c(cos 0)

c = 1

hence solution becomes

y= x + cosx