# Homework Help: Bernoulli Equation

1. Sep 19, 2010

I have solved the following Bernoulli equation by letting Z = y1 - 2:

xy' - 2y = x3y2. I obtained the solution

y = 5/(5c1 - x5)

which Wolfram Alpha has confirmed.

From this result, I have obtained y' to be

y' = 25x4/(5c1 - x5)2

The problem is when I go to check the solution by plugging into DE:

x*25x4/(5c1 - x5)2 - 2*5/(5c1 - x5)

$$= \frac{25x^4}{(5c_1 - x^5)^2} - \frac{10}{5c_1 - x^5} = \frac{25x^4}{(5c_1 - x^5)^2} - \frac{10*(5c_1 - x^5)}{(5c_1 - x^5)^2}$$

which will never equal the right hand side of the original equation:

$$x^3y^2 = x^3\frac{25}{(5c_1 - x^5)^2}$$

Anyone seeing where I am messing this up?

thanks.

2. Sep 19, 2010

### ╔(σ_σ)╝

Hello fellow engineer ,

First of all your y(x) and wolframs y(x) are not the same function.

You are missing an $$x^{2}$$ term.

3. Sep 19, 2010

Ahhh....yes. Thanks! So let's try this again....

The solution to xy' - 2y = x3y2 is given by

$$y(x) = \frac{5x^2}{5c - x^5} = 5x^2(5c - x^5)^{-1}$$

$$\Rightarrow y'(x) = (5x^2)\cdot[(-1)(5c - x^5)^{-2}(-5x^4)] + ((5c - x^5)^{-1})\cdot[10x]$$

$$\Rightarrow y'(x) = \frac{25x^6}{(5c-x^5)^2}+\frac{10x}{5c-x^5}$$

$$\Rightarrow y'(x) = \frac{25x^6+(5c - x^5)(10x)}{(5c - x^5)^2}$$

Plugging the result back into original DE we have

$$\frac{25x^7+(5c - x^5)(10x^2)}{(5c - x^5)^2} - 2*\frac{5x^2}{5c - x^5}= \frac{25x^7+(5c - x^5)(10x^2)}{(5c - x^5)^2} -\frac{(5c - x^5)(10x^2)}{(5c - x^5)^2} = \frac{25x^7}{(5c - x^5)^2} = A$$

From the right side of original DE we know that

$$A' = x^3y^2 = \frac{(x^3)(25*x^4)}{(5c - x^5)^2} =\frac{25x^7}{(5c - x^5)^2}$$

Thus, A = A' and we must have the solution.

Thanks for catching that error!