Answer to Question #126304 in Differential Equations for jse

Question

1. Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation

x dy/dx - (1+x)y = x(y)^2


2. Solve the given initial-value problem. The DE is a Bernoulli equation.

x^2 dy/dx - 2xy = 4y^4, y(1)= 1/2

Answer 1

Question 1: $x\frac{dy}{dx}-(x+1)y=xy^2$

Solution: Dividing equation by $-xy^2$ :

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

Substituting $z=y^{1-2}=y^{-1}$ , $\frac{dz}{dx}=z'=-y^{-2}y'$ , therefore: (1)

$z'+\frac{x+1}{x}z=-1$

Substituting $z=uv$ , $z'=u'v+uv'$ : (2)

$u'v+uv'+\frac{x+1}{x}uv=-1$

$u'v+u(v'+\frac{x+1}{x}v)=-1$ (3)

Setting the part inside () to 0 and separating variables:

$v'+\frac{x+1}{x}v=0$

$\frac{dv}{dx}=- \frac{x+1}{x}v$

$\int\frac{dv}v = -\int\frac{x+1}{x}dx$

$ln(v)=-(x+ln(x)+ln(k))$

$v=e^{-(x+ln(x))}=\frac1 {kxe^x}$ (4)

Putting $v$ from (4) into equation (3):

$\frac1 {kxe^x}u'=-1$

$\int du=-k\int xe^x$

$u=-k(xe^x-e^x+e^x+C)=-k(e^x(x-1)+C)$

Finding z from (2):

$z=uv=\frac{-(e^x(x-1)+C)}{xe^x}$

Finding y from (1):

$y=z^{-1}=-\frac{xe^x}{e^x(x-1)+C}$

Answer 1: $y=-\frac{xe^x}{e^x(x-1)+C}$

Question 2: $x^2y'-2xy=4y^4$ , $y(1)=\frac 12$

Solution: Dividing equation by $-\frac1 3 x^2 y^4$ :

$-3y^{-4}y'+\frac 6 x y^{-3}=-\frac {12} {x^2}$

Substituting $z=y^{-3}$ , $z'=-3y^{-4}y'$, therefore: (1)

$z'+\frac6xz=-\frac{12}{x^2}$

Substituting $z=uv , z'=u'v+uv'$ : (2)

$u'v+uv'+\frac6xuv=-\frac{12}{x^2}$

$u'v+u(v'+\frac6xv)=-\frac{12}{x^2}$ (3)

Setting the part inside () to 0 and separating variables:

$v'+\frac6xv=0$ (4)

$v'=-\frac6xv$

$\int \frac {dv} v = -6 \int \frac {dx} x$

$ln(v)= -6 ln(x)-ln(k)=ln(kx^{-6})$

$v=\frac 1 {kx^6}$

Putting $v$ from (4) into equation (3):

$\frac 1 {kx^6}u'=-\frac{12}{x^2}$

$\int du = \int -12kx^4dx$

$u= k(-\frac {12x^5}5+C)$

Finding z from (2):

$z=uv=-(\frac {12x^5}5+C)*\frac1{x^6}$

Finding y from (1):

$y=\sqrt[3]{\frac{5x^6}{12x^5+5C}}$

Assuming ​ $y(1)=\frac 12$ : $\sqrt[3]{\frac{5*(1)^6}{12*(1)^5+5C}}=\frac12$

$\sqrt[3]{\frac{5}{12+5C}}=\frac12$

$\frac{5}{12+5C}=\frac18$

$C=\frac{52}5$

Substitute C into general solution:

$y=\sqrt[3]{\frac{5x^6}{12x^5+52}}$

Answer 2: $y=\sqrt[3]{\frac{5x^6}{12x^5+52}}$