Solve the following Bernoulli differential equations b ytan

Solve the following Bernoulli differential equations

b. y\'-(tan x)y = (cos x)y, y(0) = 3.

Solution

(b)

y\' -tanx y = cosx y^4

let v = y^(-3)... => v\' = (-3)y^(-4)y\' ......=>v(0) = 1/27

=>

-v\'/3 -tanx *v = cosx

=>

v\' +(3tanx)v = -3cosx

mulplying the equation with sec^3 x, we get

v\'sec^3 x + 3vsinx/cos^4 x = -3/cos^2 x

=>

d(v/cos^3 x) = -3dx/cos^2 x

=>

v/cos^3x = -3tanx +c

=>

v = -3sinx cos^2 x +c

=>

c=1/27 since v(0) = 1/27

=>

v = -3sinx cos^2 x +1/27

=>

y^3 = 27/(1-81sinxcos^2 x)

(c)

y\' = xy^3 -y

let v = y^(-2) ..=> v\' = (-2)y^(-3)y\'.... and v(0) = 1

=>

-v\'/2 + v = x

=>

v\'-2v = -2x

=>

e^(-2x) *v\' -2v* e^(-2x) = -2xe^(-2x)

=>

d(ve^(-2x)) = -2xe^(-2x)dx

=>

ve^(-2x) = (1+2x)/2 * e^(-2x) +c

=>

1 = c+1/2 since v(0) =1

=>

c = 1/2

=>

ve^(-2x) = (1+2x)/2 * e^(-2x) +1/2

=>

v = (2x+1)/2 + e^(2x)/ 2

=>

v = (e^2x + 2x +1 )/2

=>

1/y^2 = (e^2x + 2x +1 )/2

=>

y^2 = 2/(e^2x + 2x +1 )