Question

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

Answer 1

Answer:

y= x + cosx

Step-by-step 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. = $e^{\int tanx dx} = e^{ln\ secx} = secx$

$y=\frac{1}{I.F}(\int Q dx + c)$

$y=\dfrac{1}{secx}(\int secx dx + c)$

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

using  y(0) = 1

1 = 0 +  c(cos 0)

c = 1

hence solution becomes

y= x + cosx