Verify that y=sin xcos x-cos x is a solution of the initial value problem y'+(tan x)y=cos^2 xy(0)=-1 on the interval -pi /2 < x< pi/2

Question

Verify that

y = \sin x \cos x - \cos x

is a solution of the initial value problem

y^{'} + (\tan x) y = c o s^{2} x y (0) = - 1

on the interval

- π / 2 < x < π / 2

Answer 1

y'+y P(x)= Q(x)

y^{'} + (\tan x) y = \cos^{2} x

linear diff. eqn.

P (x) = \tan x

Q (x) = \cos^{2} x

integration factor

i (x) = \exp (\int P (x) d x) = \exp (\int \tan (x) d x)

= \exp (- \ln (\cos x) = \frac{1}{\cos x}

y . i (x) = \int i (x) Q (x) d x + c

y * \frac{1}{\cos x} = \int \frac{1}{\cos x} \cos^{2} x d x + c

y * \frac{1}{\cos x} = \sin x + c

y = \sin x . \cos x + c . \cos x

y (0) = - 1 \Rightarrow - 1 = \sin 0 \cos 0 + c . c o s 0 \Rightarrow c = - 1

y = \sin x . \cos x - \cos x

Answers (1)