Mathematics, 08.10.2019 22:20 gabe2111

Abernoulli differential equation is one of the form(dy/dx)=p(x)y=q(x)y^n (*)observe that, if n=0 or 1, the bernoulli equation is linear. for other values of n, the substitution u=y^(1-n) transforms the bernoulli equation into the linear equation(du/dx)+(1-n)p(x)u=(1-n)q(x )use the appropriate substituion to solve the equationxy'+y=3xy^2and find the solution that satisfies y(1)=-8

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Answer from: drippyc334

first part: y = (1)/(x(-3Ln|x|+ C)).

Second part with y(1)=-8; y= (1)/(x(-3Ln|x|- (1/8)))

Step-by-step explanation:

Considering the given substitution in the exercise and solving some algebraic operations, the differential equation can be easily solved, as follows;

Diving the expression $xy'+y=3xy^{2}$ by x, the equation is simplified and it is possible to see the Bernoulli's equation. Thus,

$\frac{(xy'+y)}{x}=\frac{3xy^{2}}{x}\\\\\\y'=3y^{2}-\frac{y}{x}$

This Bernoulli's equation has some criteria to be solved and fortunately in the exercise is written how. In that manner, n represents the exponent of the independent variable. P(x) and Q(x) are functions and they are written as coefficients. Such algebraic criteria is more convenient to solve the differential equation, thus;

$n=2\\Q(x)=3\\P(x)=\frac{1}{x}\\ u=y^{1-n}=y^{-1}$

What does u stand for?

"u" stands for a substitution that facilitates to do the differential equation. With the given substitutions, it proceeds to rewrite the equation in terms of "u". (Bear in mind that this substitution is exactly the same given in the exercise statement)

$\frac{1}{(1-2)}u' +\frac{u}{x}=3$

Multiplying by -1 both sides of the above equation and using the integrating factor method, the differential equation is written as,

$(-1)(-u'+\frac{u}{x})=(-1)(3)\\\\u^{'}-\frac{u}{x}= -3 \\\\e^{\int(-\frac{1}{x})dx}u^{'}-\frac{u}{x}e^{\int(-\frac{1}{x})dx}= -3e^{\int(-\frac{1}{x})dx}\\\\\frac{u^{'}}{x}-\frac{u}{x^{2}}=-\frac{3}{x}\\$

The last term corresponds to a resultant derivative of a product, then

$\frac{u^{'}}{x}-\frac{u}{x^{2}}=-\frac{3}{x}\\\\\frac{d}{dx}\frac{u}{x}=-\frac{3}{x}\\\\\frac{du}{x}=-3\int(\frac{1}{x})dx\\\\u=-3xln|x|+xc_{1}$

As "u" represents a substitution, it is necessary to rewrite the answer is terms of "y", then

$u=-3xln|x|+xc_{1}\\\\\frac{1}{y}=-3xln|x|+xc_{1}\\\\y=\frac{1}{x(-3ln|x|+c_{1})}$

Now, considering the initial condition in which y(1)=-8; thus

$y=\frac{1}{x(-3ln|x|+c_{1})}\\\\-8=y=\frac{1}{1(-3ln|1|+c_{1})}\\\\-8=\frac{1}{c_{1}}\\\\c_{1}=-\frac{1}{8}$

The equation is therefore

$y=\frac{1}{x(-3ln|x|-\frac{1}{8})}$

For this exercise was employed an algebraic substitution (Bernoulli's equation) and a common method of solving differential equations (integrating factor). Before to proceed the use of the given substitutions, another methods were considered, nevertheless those methods did not allow the separation of differential equations for solving it easier.

Answer from: Quest

It took off from 11 degrees

Answer from: Quest

the graph is negative for all real values of x where -6 < x < -2.

step-by-step explanation:

this refers to the values of f(x) in the area below the x-axis.

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