Question

Solve the following differential equation: 2dy/dx + y = y^3(x - 1)

Answer 1

$y'+y/2=(xy^3)/2-y^3/2$

Bernuli equalation

$y'+a(x)y=b(x)y^n$

a(x)=1/2

b(x)=(x-1)/2

n=3

$y'/y^3+1/(2y^2)=x/2-1/2$

Deleting, we miss solution y=0

$u=1/y^2$

$u'=-2y'/y^3$

$y=1/\sqrt u$

$y'=-u'y^3/2$

$u/2-u'/2=x/2-1/2$

u-u'=x-1

-u'=x-u-1

v=x-u-1

v'=1-u'

u=x-v-1

u'=1-v'

v'-1=v

v'=v+1=dv/dx

dv=(v+1)dx

dv/(v+1)=dx

Deleting, we miss solution v+1=0, x=1/y²

$\int dv/(v+1)=\int dx$

ln (v+1)=x+C

v+1=e^x+C

x-u=Ce^x

x-1/y²=Ce^x

$y=\sqrt {\frac{1}{x-Ce^x}}$