Question

Bernoulli's Equation diff equation

Answer 1

I cant reduce this to the standard form of bernoulli.. help

Answer 2

\[\begin{align*}dy+y\,dx&=2xy^2e^x\,dx\end{align*}~~?\]

Answer 3

yes yes

Answer 4

First divide through by the differential of \(x\) and the highest power of \(y\),
\[\begin{align*}y^{-2}\frac{dy}{dx}+y^{-1}&=2xe^x\end{align*}\]
For the Bernoulli substitution, you would consider \(t=y^{-1}\), which gives \(\dfrac{dt}{dx}=-y^{-2}\dfrac{dy}{dx}\).
\[\begin{align*}-\frac{dt}{dx}+t&=2xe^x\end{align*}\]
and now you have an ODE linear in \(t\).

Answer 5

my integrating factor would be e^ t^2/2?

Answer 6

what would be my n?

Answer 7

Not quite, it's much simpler than that.
\[\frac{dt}{dx}-t=-2xe^x\]
which would require an IF of
\[\mu(x)=\exp\left(-\int\,dx\right)=e^{-x}\]

Answer 8

yes.. im really confused with the n.

Answer 9

i dont have t on the right side

Answer 10

i mean y

Answer 11

I'm not sure what you mean by "the n." I think you're referring to the exponent of the nonlinear term... A Bernoulli eq. has the general form
\[\frac{dy}{dx}+f(x)y=g(x)y^n\]
where \(n\not=0,1\) (otherwise you have a linear ODE). Is that the \(n\) you're talking about?

Answer 12

ah yes!

Answer 13

omg the first reminder in bernoulli.. Thank you so much

Answer 14

and im confused with my integrating factor

Answer 15

When you have a linear equation (i.e. of the form \(y'+f(x)y=g(x)\)) the integrating factor is
\[\ln\mu(x)=\int f(x)\,dx~~\iff~~\mu(x)=\exp\left(\int f(x)\,dx\right)\]
The Bernoulli substitution changes the variable so that a nonlinear ODE can be reduced to a linear one.

Answer 16

yes yes but my f(x) is t or y^-1

Answer 17

Given ODE (with the differential divided through):
\[\frac{dy}{dx}+\underbrace{1}_{\large f(x)}\,y=\underbrace{2xe^x}_{\large g(x)}\,y^2\]
(The assignments of \(f\) and \(g\) here apply to this Bernoulli form only.)

When you carry out the substitution, you get the linear form
\[\frac{dt}{dx}-\underbrace{1}_{\large f(x)}t=\underbrace{-2xe^x}_{\large g(x)}\]
(Now \(f\) and \(g\) are different functions in this linear form - they are distinct from those in the Bernoulli form.)

Answer 18

i think my integrating factor would be e^x

Answer 19

The takeaway here is that you have to keep track of which variable is dependent, especially AFTER you make any substitution.