New Notes Blog: `xy' + y = xy^3` Solve the Bernoulli differential equation.

Monday, 15 September 2014

$\displaystyle{x}{y}'+{y}={x}{{y}}^{{3}}$ Solve the Bernoulli differential equation.

Given equation is $\displaystyle{x}{y}'+{y}={x}{{y}}^{{3}}$

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

is called as the Bernoullis 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 $\displaystyle{u}={{y}}^{{{1}-{n}}}$

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

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

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...

Given equation is $\displaystyle{x}{y}'+{y}={x}{{y}}^{{3}}$

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

is called as the Bernoullis 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 $\displaystyle{u}={{y}}^{{{1}-{n}}}$

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

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

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 ,

$\displaystyle{x}{y}'+{y}={x}{{y}}^{{3}}$

=> y'+(1/x)y =y^3

and

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

on comparing we get

$\displaystyle{P}={\left(\frac{{1}}{{x}}\right)}{Q}={1},{n}={3}$

so the linear form of first order of the equation $\displaystyle{x}{y}'+{y}={x}{{y}}^{{3}}$ is given as

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

=> $\displaystyle\frac{{{u}'}}{{{1}-{3}}}+{\left(\frac{{1}}{{x}}\right)}{u}={1}$

=> $\displaystyle-\frac{{{u}'}}{{2}}+\frac{{u}}{{x}}={1}$

=> $\displaystyle{\left({u}'\right)}-{2}\frac{{u}}{{x}}=-{2}$

so this linear equation is of the form

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

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

so I.F (integrating factor ) = $\displaystyle{{e}}^{{\int{p}{\left.{d}{x}\right.}}}={{e}}^{{\int{\left(-\frac{{2}}{{x}}\right)}{\left.{d}{x}\right.}}}={{e}}^{{{\ln{{\left(\frac{{1}}{{{x}}^{{2}}}\right)}}}}}=\frac{{1}}{{{x}}^{{2}}}$

and the general solution is given as

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

=> $\displaystyle{u}{\left(\frac{{1}}{{{x}}^{{2}}}\right)}=\int{\left(-{2}\right)}\cdot{\left(\frac{{1}}{{{x}}^{{2}}}\right)}{\left.{d}{x}\right.}+{c}$

=> $\displaystyle{u}{\left(\frac{{1}}{{{x}}^{{2}}}\right)}={\left(-{2}\right)}\int{\left(\frac{{1}}{{{x}}^{{2}}}\right)}{\left.{d}{x}\right.}+{c}$

so now,

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

=> $\displaystyle{u}{\left(\frac{{1}}{{{x}}^{{2}}}\right)}={\left({2}\right)}{\left({{x}}^{{-{{1}}}}\right)}+{c}$

=> $\displaystyle{u}{\left(\frac{{1}}{{{x}}^{{2}}}\right)}={\left(\frac{{2}}{{x}}\right)}+{c}$

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

but $\displaystyle{u}={{y}}^{{-{{2}}}}$

so,

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

=> $\displaystyle{{y}}^{{2}}=\frac{{1}}{{{\left({2}{x}\right)}+{c}\cdot{{x}}^{{2}}}}$

=> $\displaystyle{y}=\sqrt{{\frac{{1}}{{{\left({2}{x}\right)}+{c}\cdot{{x}}^{{2}}}}}}$

=> $\displaystyle{y}=\sqrt{{\frac{{1}}{{{2}{x}+{c}{{x}}^{{2}}}}}}$

the general solution.

New Notes Blog

Monday, 15 September 2014

$\displaystyle{x}{y}'+{y}={x}{{y}}^{{3}}$ 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?

Monday, 15 September 2014

Solve the Bernoulli differential equation.

No comments:

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?

$\displaystyle{x}{y}'+{y}={x}{{y}}^{{3}}$ 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?