Answer to Question #126305 in Differential Equations for jse

Question

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

dy/dx = y(xy^6 - 1)

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

x dy/dx + y = 1/y^2

Answer 1

$$Solution$$

1. To solve the DE Bernoulli equation,

$dy/dx = y(xy^6 - 1)$

The standard form is

$\frac{dy}{dx}+P(x)y=f(x)y^n$

$Where$ $n$ $is$ $not$ $equal$ $to$ $zero$ $or$ $one$

$dy/dx = xy^7 - y$

$dy/dx+y = xy^7$ $-------------->(1)$

$n=7$

$U=y^{1-n}=y^{1-7}=y^{-6}$

$U=y^{-6}$

$U^{-\frac{1}{6}}=(y^{-6})^{-\frac{1}{6}}$ $\implies$ $y=U^{-\frac{1}{6}}$

$dy/dx =- \frac{1}{6}U^{- \frac{7}{6}}*\frac{du}{dx}$

$\frac{1}{6}U^{- \frac{7}{6}}\frac{du}{dx}+U^{-\frac{1}{6}}=x U^{- \frac{7}{6}}----------->(2)$

Multiply by $( \frac{1}{6}U^{- \frac{7}{6}})$

$\frac{du}{dx}-6U^1=-6x$

$y(x)=\epsilon^{\int \rho (x)dx}=\epsilon^{\int -6dx}= \epsilon^{-6x}$

$\implies y(x)=\epsilon^{-6x}$

$\epsilon^{-6x} \frac{du}{dx}-6 \epsilon^{-6x} U=-6x\epsilon^{-6x}$

$\frac{d}{dx}[\epsilon^{-6x} U]=-6x\epsilon^{-6x}$

Integration:

$\epsilon^{-6x} U=\int -6x\epsilon^{-6x} dx$

$\epsilon^{-6x} U=x\epsilon^{-6x} +\frac{1}{6}\epsilon^{-6x} +C$

$\implies U=x+\frac{1}{6}+\frac{C}{\epsilon^{-6x} }$

$y^{-6}=x+\frac{1}{6}+C\epsilon^{6x} ------------->Answer$

2. To solve the DE Bernoulli equation.

$x \frac{dy}{dx} + y = \frac{1}{y^2}$

The standard form is

$\frac{dy}{dx}+P(x)y=f(x)y^n$

Where n is not equal to zero or one

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

$n=-2$

$U=y^{1-n}=y^{1-(-2)}=y^3$

$U=y^{3}$

$\implies y=U^{\frac{1}{3}}$

$\frac{dy}{dx}=\frac{1}{3}U^{-\frac{2}{3}}\frac{du}{dx}$

$\frac{1}{3}U^{-\frac{2}{3}}\frac{du}{dx}+\frac{1}{x}U^{\frac{1}{3}}=\frac{1}{x}U^{-\frac{2}{3}}------->(2)$

Multiply by $(3U^{-\frac{2}{3}})$

$\frac{du}{dx}+\frac{3}{x}U^1=\frac{3}{x}$

$y(x)=\epsilon^{\int \rho (x)dx}=\epsilon^{\int \frac{3}{x}dx}= \epsilon^{3ln|x|}=|x^3|=x^3$

$\implies y(x)=x^3$

$x^3\frac{du}{dx}+3x^2U=3x^2$

$\frac{d}{dx}[x^3U]=3x^2$

$x^3U=\frac{3x^2}{3}+C$

$U=1+\frac{C}{x^3}$

$y^3=1+\frac{C}{x^x} \implies y=(1+\frac{C}{x^x})^{\frac{1}{3}}------->Answer$