# Bernoulli differential equation, i got the form down, where did i f it up? :-x

1. Jan 22, 2006

### mr_coffee

Hello everyone! ITS ME! i'm having a good time with some Bernoulli differential equations, and yet it didn't work. Here is the directions:
A Bernoulli differential equation is one of the form:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/c4/3f85cf0e6820be855c6d2a21d051b71.png [Broken]
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
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/aa/f6e0ed9c0e920a9374d9b874f69ff21.png [Broken]
Use an appropriate substitution to solve the equation
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/a5/7eb1214eb5ca480f1d8188f6dd4e9a1.png [Broken]
and find the solution that satisfies y(1)=3.

y(x) =

Okay here is my work:
P(x) = 1/x;
Q(x) = 9
n = 2

u' + (-1)(1/x)*u = (-1)*9
so integrating facotr would be:
I(t) = e^(-1/x)
I = -x;

-x*u' + u = 9x;
integration gives:
u = (-9/2)*x -c/x

applying intial condition: y(1) = 3;
3 = -9/2 - C/1
C = -15/2;

Plugging in constant of integration and solving for u gives me:
u = (-9/2)*x +(15/2);
which is wrong of course! any idea why? Thank you! (this is the first time doing a problem of this form so im' not sure if i'm suppose to plug somthing back in or what?);

THANKS

Last edited by a moderator: May 2, 2017
2. Jan 22, 2006

### TD

From a first look, your Q(x) is wrong. As you see, Q may be a function of x and you just took 9, but the coefficient of y² was 9x.

By using the Bernouilli substitution, i.e. z = 1/y, your DE becomes

$$xy' + y = 9xy^2 \mathop \to \limits_{z' = - y'/y^2 }^{z = y^{ - 1} } - xz' + z = 9x$$

3. Jan 23, 2006

### mr_coffee

Thanks TD, i'm kinda confusd on your notation of
z = y^-1
->
z' = -y'/y^2

what do u mean? Also I fallowed an example program my professor did and he just used the number, but i agree i think it should be 9x

4. Jan 23, 2006

### TD

Well my z is just your u, for the Bernouilli equation of degree n, the substitution is - as you said - u = y^(1-n). With n being 2 here, and your u being my z, that gives us z = y^(-1), no? Then z' is just dz/dx.