Question

solve the differential equation using Bernoulli
y'-(3/x)y=(y^3/X^5) Find the solution that satisfies y(1)=1

Answer 1

\[y'-\frac{ 3 }{ x }y=\frac{ y^3 }{ x^5 }\]

Answer 2

\[y'-\frac{3}{x}y=\frac{y^3}{x^5}~~\Rightarrow~~y^{-3}y'-\frac{3}{x}y^{-2}=\frac{1}{x^5}\]
Substitute: \(u=y^{-2}\) so that \(u'=-2y^{-3}y'\):
\[-\frac{1}{2}u'-\frac{3}{x}u=\frac{1}{x^5}\\
u'+\frac{6}{x}u=-\frac{2}{x^5}\]

Answer 3

Do you know how to solve linear ODE's? Find the integrating factor, etc?

Answer 4

I have a problem with linear for some reason.. I get my integrating factor to be x^6 so
\[u(x)= \frac{ 1 }{ x^6 }(\int\limits(x^6\frac{ -2 }{ x^5 }dx+c)\]

Answer 5

Right on with that integrating factor:\[\mu(x)=x^6\]
Equation becomes\[x^6u'+6x^5u=-2x\\
\frac{d}{dx}\left[x^6u\right]=-2x\\
x^6u=-2\int x~dx\\
x^6u=-x^2+C\\
u=-\frac{1}{x^4}+\frac{C}{x^6}\]

Answer 6

how did you get the second line of that?

Answer 7

the formula I have is \[y(x) = \frac{ 1 }{u }(\int\limits u*p(x)*q(x) dx +c)\]

Answer 8

\[y(x) = x ^{-6}(-x ^{2}+c)^{-1/2}\] after plugging in the u to the original

Answer 9

\[\color{red}{x^6u'+6x^5u}=-2x\]
the red part is just a derivative. Use the "reverse" product rule, since
\[\frac{d}{dx}\left[x^6u\right]=x^6u'+6x^5u\]

Answer 10

i cant read that at all