solve y′−6y/x=y^5/x^13
i have tried again and again but my answer is still wrong. can someone show me the steps to solving this question?
My attempt to solve it failed. I'm new to differentual equations. I saw this problem and thought that is easier than my homework. It turns out that I need help with this one, too. I don't know how you tried, but this is where things went wrong for me.
It was not that obvious to me that this is a Bernoulli equation $y^′−\frac{6y}{x}=\frac{y^5}{x^{13}}$, so I did some rearrangements. $y^′=\frac{y^5}{x^{13}}+\frac{6y}{x}$ and $y^′=6yx^{-1}+y^5x^{-13}$ now I see it clearly.
The first step from the algorithm I was taugh is to solve for the linear term.
$\frac{\text{d}\bar{y}}{\text{d}x} = 6\bar{y}x^{-1}$
$\bar{y}= Ce^{\int_{}^{}x^{-1}dx } = Ce^{ln(x)} = Cx$
DONE.
Now I'm looking for solutions of this form. $\bar{y}(x)=C(x)x$
We have: $C(x)^′x+C(x)= 6C(x)+x^{-8}C(x)^5$ (if I'm not wrong)
My teacher said that the e-terms ( C(x) and 6C(x) ) should always cancel each other, otherwise you made mistakes. I think I'm wrong, but I don't know where.
Take a look here: I need some help with this bernoulli's equation that is given in my assignment
I think Hippasus is confusing Bernoulli equations with linear equations.
I learned an algorithm at university. The first step of solving a Bernoulli equation is to consider the linear equation associeted to it.
Bernoulli's $\frac{\text{d}x}{\text{d}t}=a(t)x+b(t)x^\alpha, \alpha \in \Re\diagdown \left\{1,0\right\}$
Linear $\frac{\text{d}x}{\text{d}t}=a(t)x$
Am I mistaking ?