Welcome to our community

Be a part of something great, join today!

Solving the Bernoulli D.E

bergausstein

Active member
Jul 30, 2013
191
Solve the ff:

$\displaystyle\frac{dy}{dx}-y=xy^5$

$\displaystyle\frac{dy}{dx}-\frac{y}{x}=-\frac{5}{2}x^2y^3$

can you help start solving these problems? thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
These are Bernoulli equations. Such an equation may be written in the form:

\(\displaystyle \frac{dy}{dx}+P(x)y=Q(x)y^n\)

Use of the substitution:

\(\displaystyle v=y^{1-n}\)

will transform the ODE into a linear equation.
 

bergausstein

Active member
Jul 30, 2013
191

the n=5

$\displaystyle v=y^{1-5}=\frac{1}{y^4}$

$\displaystyle y=(\frac{1}{v})^{\frac{1}{4}}$

$\displaystyle y'=-\frac{1}{4}v^{-\frac{5}{4}}\frac{dv}{dx}$

substitute,

$\displaystyle y=(\frac{1}{v})^{\frac{1}{4}}$ and $\displaystyle y'=-\frac{1}{4}v^{-\frac{5}{4}}\frac{dv}{dx}$ to the original D.E

$\displaystyle -\frac{1}{4}v^{-\frac{5}{4}}\frac{dv}{dx}-\frac{1}{v^{\frac{1}{4}}}=x\frac{1}{v^{\frac{5}{4}}}$

multiplying both sides by $v^{\frac{5}{$}}$ and -4.

$\displaystyle \frac{dv}{dx}+4v=-4x$

the integrating factor is $e^{4x}$

$\displaystyle e^{4x}\frac{dv}{dx}+4ve^{4x}=-4xe^{4x}$

$\displaystyle \int D_x(e^{4x}v)=\int -4xe^{4x}$ the rhs using by parts

$\displaystyle e^{4x}v= -e^{4x}+\frac{1}{4}e^{4x}+c$

dividing both sides by $ e^{4x}$

$\displaystyle v= -1+\frac{1}{4}+\frac{c}{e^{4x}}$

now since $v=\frac{1}{y^4}$

$\displaystyle\frac{1}{y^4}=\frac{c}{e^{4x}}-\frac{3}{4}$ --->> this would be my implicit general solution

is my answer correct?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We are given:

\(\displaystyle \frac{dy}{dx}-y=xy^5\)

Use the substitution:

\(\displaystyle v=y^{-4}\implies\frac{dv}{dx}= -4y^{-5}\frac{dy}{dx}\)

Divide the original ODE by$y^5$ observing we are eliminating the trivial solution $y\equiv0$. We obtain:

\(\displaystyle y^{-5}\frac{dy}{dx}-y^{-4}=x\)

Applying the substitution we obtain:

\(\displaystyle \frac{dv}{dx}+4v=-4x\)

Multiply through by $e^{4x}$:

\(\displaystyle e^{4x}\frac{dv}{dx}+4e^{4x}v=-4xe^{4x}\)

This becomes:

\(\displaystyle \frac{d}{dx}\left(e^{4x}v \right)=-4xe^{4x}\)

Integrating (using IBP on the right (which you did incorrectly)), we obtain:

\(\displaystyle e^{4x}v=\frac{1}{4}e^{4x}\left(1-4x \right)+C\)

\(\displaystyle v=\frac{1}{4}\left(1-4x \right)+Ce^{-4x}\)

\(\displaystyle y^{-4}=\frac{1}{4}\left(1-4x \right)+Ce^{-4x}\)

Thus the solution may be given implicitly by:

\(\displaystyle y^4=\frac{4}{1-4x+Ce^{-4x}}\)

Note: the trivial solution we eliminated is not obtainable by the implicit solution gave above.
 

bergausstein

Active member
Jul 30, 2013
191
can you tell me where's my mistake in my solution above?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
can you tell me where's my mistake in my solution above?
Edit: There is a small mistake in the integration. Sorry.

[tex]\displaystyle \begin{align*} \int{-4x\,e^{4x}\,dx} = -4x\,e^{4x} + \frac{1}{4}e^{4x} + C \end{align*}[/tex]
 
Last edited:

bergausstein

Active member
Jul 30, 2013
191
$\displaystyle e^{4x}v= -xe^{4x}+\frac{1}{4}e^{4x}+c$

$\displaystyle v= -x+\frac{1}{4}+\frac{c}{e^{4x}}$

my final answer should be


$\displaystyle\frac{1}{y^4}=\frac{c}{e^{4x}}-x+\frac{1}{4}$

am I right?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
$\displaystyle e^{4x}v= -xe^{4x}+\frac{1}{4}e^{4x}+c$

$\displaystyle v= -x+\frac{1}{4}+\frac{c}{e^{4x}}$

my final answer should be


$\displaystyle\frac{1}{y^4}=\frac{c}{e^{4x}}-x+\frac{1}{4}$

am I right?
Please re-read my last post. You keep missing the factor of -4 on the first term.
 

bergausstein

Active member
Jul 30, 2013
191
Please re-read my last post. You keep missing the factor of -4 on the first term.
$\displaystyle\int -4xe^{4x}dx=-4\int xe^{4x}$

let $u=x$ and $du=dx$

let $dv=e^{4x}dx$ now $v=\frac{1}{4}e^{4x}$

$uv-\int vdu$

$\frac{1}{4}xe^{4x}-\frac{1}{4}\int e^{4x}dx$

$-4\left(\frac{1}{4}xe^{4x}-\frac{1}{4}\frac{1}{4}e^{4x}+c\right)= -xe^{4x}+\frac{1}{4}e^{4x}+c$

isn't correct?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
$\displaystyle\int -4xe^{4x}dx=-4\int xe^{4x}$

let $u=x$ and $du=dx$

let $dv=e^{4x}dx$ now $v=\frac{1}{4}e^{4x}$

$uv-\int vdu$

$\frac{1}{4}xe^{4x}-\frac{1}{4}\int e^{4x}dx$

$-4\left(\frac{1}{4}xe^{4x}-\frac{1}{4}\frac{1}{4}e^{4x}+c\right)= -xe^{4x}+\frac{1}{4}e^{4x}+c$

isn't correct?
Never mind, what you have is correct. I forgot the 1/4 factor in the v term.

I think this many mistakes means it's time for bed >_<