New Notes Blog: `y' + xy = xy^-1` Solve the Bernoulli differential equation.

Tuesday, 20 January 2015

$\displaystyle{y}'+{x}{y}={x}{{y}}^{{-{{1}}}}$ Solve the Bernoulli differential equation.

Given equation is $\displaystyle{y}'+{x}{y}={x}{{y}}^{{-{1}}}$

An equation of the form $\displaystyle{y}'+{P}{y}={Q}{{y}}^{{n}}$

is called as the Bernoulli equation .

so, to proceed to solve this equation we have to transform the equation into a linear equation form of first order as follows

=> $\displaystyle{y}'{\left({{y}}^{{-{{n}}}}\right)}+{P}{{y}}^{{{1}-{n}}}={Q}$

let $\displaystyle{u}={{y}}^{{{1}-{n}}}$

=> $\displaystyle{\left({1}-{n}\right)}{{y}}^{{-{n}}}{y}'={u}'$

=> $\displaystyle{{y}}^{{-{n}}}{y}'=\frac{{{u}'}}{{{1}-{n}}}$

so ,

$\displaystyle{y}'{\left({{y}}^{{-{{n}}}}\right)}+{P}{{y}}^{{{1}-{n}}}={Q}$

=> $\displaystyle\frac{{{u}'}}{{{1}-{n}}}+{P}{u}={Q}$

so this equation is now of the linear...

Given equation is $\displaystyle{y}'+{x}{y}={x}{{y}}^{{-{1}}}$

An equation of the form $\displaystyle{y}'+{P}{y}={Q}{{y}}^{{n}}$

is called as the Bernoulli equation .

so, to proceed to solve this equation we have to transform the equation into a linear equation form of first order as follows

=> $\displaystyle{y}'{\left({{y}}^{{-{{n}}}}\right)}+{P}{{y}}^{{{1}-{n}}}={Q}$

let $\displaystyle{u}={{y}}^{{{1}-{n}}}$

=> $\displaystyle{\left({1}-{n}\right)}{{y}}^{{-{n}}}{y}'={u}'$

=> $\displaystyle{{y}}^{{-{n}}}{y}'=\frac{{{u}'}}{{{1}-{n}}}$

so ,

$\displaystyle{y}'{\left({{y}}^{{-{{n}}}}\right)}+{P}{{y}}^{{{1}-{n}}}={Q}$

=> $\displaystyle\frac{{{u}'}}{{{1}-{n}}}+{P}{u}={Q}$

so this equation is now of the linear form of first order

Now,

From this equation ,

$\displaystyle{y}'+{x}{y}={x}{{y}}^{{-{1}}}$

and

$\displaystyle{y}'+{P}{y}={Q}{{y}}^{{n}}$

on comparing we get

$\displaystyle{P}={x},{Q}={x},{n}=-{1}$

so the linear form of first order of the equation $\displaystyle{y}'+{x}{y}={x}{{y}}^{{-{1}}}$ is given as

=> $\displaystyle\frac{{{u}'}}{{{1}-{n}}}+{P}{u}={Q}$ where $\displaystyle{u}={{y}}^{{{1}-{n}}}={{y}}^{{2}}$

=> $\displaystyle\frac{{{u}'}}{{{1}-{\left(-{1}\right)}}}+{\left({x}\right)}{u}={x}$

=> $\displaystyle\frac{{{u}'}}{{2}}+{x}{u}={x}$

=> $\displaystyle{u}'+{2}{x}{u}={2}{x}$

so this linear equation is of the form

$\displaystyle{u}'+{p}{u}={q}$

$\displaystyle{p}={2}{x},{q}={2}{x}$

so I.F (integrating factor ) = $\displaystyle{{e}}^{{\int{p}{\left.{d}{x}\right.}}}={{e}}^{{\int{2}{x}{\left.{d}{x}\right.}}}={{e}}^{{2}}\frac{{{{x}}^{{2}}}}{{2}}={{e}}^{{{{x}}^{{2}}}}$

and the general solution is given as

$\displaystyle{u}{\left({I}.{F}\right)}=\int{q}\cdot{\left({I}.{F}\right)}{\left.{d}{x}\right.}+{c}$

=> $\displaystyle{u}{\left({{e}}^{{{{x}}^{{2}}}}\right)}=\int{\left({2}{x}\right)}\cdot{\left({{e}}^{{{{x}}^{{2}}}}\right)}{\left.{d}{x}\right.}+{c}$

=> $\displaystyle{u}{\left({{e}}^{{{{x}}^{{2}}}}\right)}=\int{\left({{e}}^{{{{x}}^{{2}}}}\right)}{2}{x}{\left.{d}{x}\right.}+{c}$

let us first solve

$\displaystyle\int{{e}}^{{{{x}}^{{2}}}}{2}{x}{\left.{d}{x}\right.}$

so , let $\displaystyle{t}={{x}}^{{2}}$

$\displaystyle{\left.{d}{t}\right.}={2}{x}{\left.{d}{x}\right.}$

$\displaystyle\int{{e}}^{{{{x}}^{{2}}}}{2}{x}{\left.{d}{x}\right.}=\int{{e}}^{{{t}}}{\left.{d}{t}\right.}={{e}}^{{t}}={{e}}^{{{{x}}^{{2}}}}$

so now

=> $\displaystyle{u}{{e}}^{{{{x}}^{{2}}}}={{e}}^{{{{x}}^{{2}}}}+{c}$

=> $\displaystyle{u}=\frac{{{\left({{e}}^{{{{x}}^{{2}}}}\right)}+{c}}}{{{{e}}^{{{{x}}^{{2}}}}}}$

= $\displaystyle{1}+{c}{{e}}^{{-{{x}}^{{2}}}}$

but

$\displaystyle{u}={{y}}^{{2}}$ ,so

$\displaystyle{{y}}^{{2}}={\left({1}+{c}{{e}}^{{-{{x}}^{{2}}}}\right)}$

$\displaystyle{y}=\sqrt{{{1}+{c}{{e}}^{{-{{x}}^{{2}}}}}}$

is the general solution.

New Notes Blog

Tuesday, 20 January 2015

$\displaystyle{y}'+{x}{y}={x}{{y}}^{{-{{1}}}}$ Solve the Bernoulli differential equation.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Tuesday, 20 January 2015

Solve the Bernoulli differential equation.

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{y}'+{x}{y}={x}{{y}}^{{-{{1}}}}$ Solve the Bernoulli differential equation.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?