Question

16. Integrating factor of the differential equation
cos xdy/dx+ y sinx=1 is
(1) cosx
2) tanx
(3) secx
(4) sinx​

Answer 1

Answer:

$\red{Question} \implies \\ integreating \: factor \: of \: the \: differential \: equation \\ \\ cos \: x \frac{dy}{dx} + y \: sin \: x = 1 \: \: \\ \\ \red{solution}\implies \\ \\ \green{step \: by \: step \: solution} \\ \implies \: cos \: x \frac{dy}{dx} + y \: sin \: x = 1 \\ \\ divided \: by \: cos \: x \: on \: both \: sides \\ \\ \implies \: \red{\frac{dy}{dx} + y \frac{sin \: x}{cos \: x} = \frac{1}{cos \: x} } \\ \\ \implies \: \frac{dy}{dx} + y \: tan \: x = sec \: x \: \\ compare \: this \: equation \: by \\ \\ \implies \pink{\frac{dy}{dx} + yp = q } \\ \\ we \: get \: \\ p = tan \: x \: \: \: \: and \: \: \: \: q = sec \: x \\ \\ now \: we \: know \: that \: \\ \\ \green{integreating \: factor (i.f)= {e}^{ \int\: p.dx}} \\ \\ \implies \: i.f = {e}^{ \int \: tan \: x.dx} \\ \\ \: we \: know \: that \\ \\ \int \: tan \: x \: .dx = - log \: cos \: x \: \\ \\ therefore \: \\ \\ \implies \: i.f = {e}^{ - log \: cos \: x + c} \\ \\ \implies \: i.f = \frac{1}{cos \: x} \\ \\ \implies \: \red{\boxed{i.f = sec \: x}} \: \\ \\ hope \: it \: helps \: u$