Question #282114

1. Solve the following Bernoulli's Differential Equations. Show your solutions.



a. dy/dx + (1/3) y = e^x y²



b. x (dy/dx) + y = xy³



c. dy/dx + (2/x) y = -x² cos x y²



d. x²y-x³ (dy/dx) = y² cos x



1
Expert's answer
2021-12-28T07:18:03-0500

a.

yy2+13y=ex\frac{y'}{y^2}+\frac{1}{3y}=e^x


z=1/y,z=y/y2z=1/y,z'=-y'/y^2


z+z/3=ex-z'+z/3=e^x

z=uv,z=uv+uvz=uv,z'=u'v+uv'

(uv+uv)+uv/3=ex-(u'v+uv')+uv/3=e^x

uv+u(v/3v)=ex-u'v+u(v/3-v')=e^x


v/3v=0v/3-v'=0

uv=ex-u'v=e^x


dv/v=dx/3dv/v=dx/3

lnv=x/3lnv=x/3

v=ex/3v=e^{x/3}


uex/3=ex-u'e^{x/3}=e^x

u=e2x/3u'=-e^{2x/3}

u=3e2x/3/2+cu=-3e^{2x/3}/2+c


z=ex/3(3e2x/3/2+c)z=e^{x/3}(-3e^{2x/3}/2+c)


y=1ex/3(3e2x/3/2+c)y=\frac{1}{e^{x/3}(-3e^{2x/3}/2+c)}


b.

xyy3+1y2=x\frac{xy'}{y^3}+\frac{1}{y^2}=x


z=1/y2,z=2y/y3z=1/y^2,z'=-2y'/y^3


xz/2+z=x-xz'/2+z=x

z=uv,z=uv+uvz=uv,z'=u'v+uv'

x(uv+uv)+2uv=2x-x(u'v+uv')+2uv=2x

xvu+u(2vxv)=2x-xvu'+u(2v-xv')=2x


vu=2-vu'=2

2vxv=02v-xv'=0


dv/v=2dx/xdv/v=2dx/x

lnv=2lnxlnv=2lnx

v=x2v=x^2


x2u=2-x^2u'=2

du=2dx/x2du=-2dx/x^2

u=2/x+cu=2/x+c

z=x2(2/x+c)z=x^2(2/x+c)


y=1x2/x+cy=\frac{1}{x\sqrt{2/x+c}}


c.

y/y2+2/(xy)=x2cosxy'/y^2+2/(xy)=-x^2cosx

z=1/y,z=y/y2z=1/y,z'=-y'/y^2

z+2z/x=x2cosx-z'+2z/x=-x^2cosx

z=uv,z=uv+uvz=uv,z'=u'v+uv'

(uv+uv)+2uv/x=x2cosx-(u'v+uv')+2uv/x=-x^2cosx


uv=x2cosx-u'v=-x^2cosx

2v/xv=02v/x-v'=0


dv/v=2dx/xdv/v=2dx/x

v=x2v=x^2


u=cosxu'=cosx

u=sinx+cu=sinx+c

z=x2(sinx+c)z=x^2(sinx+c)


y=1x2(sinx+c)y=\frac{1}{x^2(sinx+c)}


d.

x2/yx3y/y2=cosxx^2/y-x^3y'/y^2=cosx

z=1/y,z=y/y2z=1/y,z'=-y'/y^2

zx2+x3z=cosxzx^2+x^3z'=cosx

z=uv,z=uv+uvz=uv,z'=u'v+uv'

uvx2+x3(uv+uv)=cosxuvx^2+x^3(u'v+uv')=cosx


x3uv=cosxx^3u'v=cosx

vx2+x3v=0vx^2+x^3v'=0


dv/v=dx/xdv/v=-dx/x

lnv=lnxlnv=-lnx

v=1/xv=1/x


x2u=cosxx^2u'=cosx


u=cosxdx/x2=i(Γ(1,ix))Γ(1,ix)2+cu=\int cosxdx/x^2=-\frac{i(\Gamma(-1,ix))-\Gamma(-1,-ix)}{2}+c


z=1x(i(Γ(1,ix))Γ(1,ix)2+c)z=\frac{1}{x}(-\frac{i(\Gamma(-1,ix))-\Gamma(-1,-ix)}{2}+c)


y=xi(Γ(1,ix))Γ(1,ix)2+cy=\frac{x}{-\frac{i(\Gamma(-1,ix))-\Gamma(-1,-ix)}{2}+c}


where Γ(z)\Gamma(z) is gamma function:

Γ(z)=0xz1exdx\Gamma(z)=\int^{\infin}_0 x^{z-1}e^{-x}dx


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