Answer to Question #282114 in Differential Equations for John

Question

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

Answer 1

a.

$\frac{y'}{y^2}+\frac{1}{3y}=e^x$

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

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

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

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

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

$v/3-v'=0$

$-u'v=e^x$

$dv/v=dx/3$

$lnv=x/3$

$v=e^{x/3}$

$-u'e^{x/3}=e^x$

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

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

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

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

b.

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

$z=1/y^2,z'=-2y'/y^3$

$-xz'/2+z=x$

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

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

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

$-vu'=2$

$2v-xv'=0$

$dv/v=2dx/x$

$lnv=2lnx$

$v=x^2$

$-x^2u'=2$

$du=-2dx/x^2$

$u=2/x+c$

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

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

c.

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

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

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

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

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

$-u'v=-x^2cosx$

$2v/x-v'=0$

$dv/v=2dx/x$

$v=x^2$

$u'=cosx$

$u=sinx+c$

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

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

d.

$x^2/y-x^3y'/y^2=cosx$

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

$zx^2+x^3z'=cosx$

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

$uvx^2+x^3(u'v+uv')=cosx$

$x^3u'v=cosx$

$vx^2+x^3v'=0$

$dv/v=-dx/x$

$lnv=-lnx$

$v=1/x$

$x^2u'=cosx$

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

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

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

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

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