Question

solve by using Bernoulli's equation:
xy'+y=-6xy^2

Answer 1

\[\begin{align*}
xy'+y&=-6xy^2\\\\
y^{-2}y'+\frac{1}{x}y^{-1}&=-6\\\\
-v'+\frac{1}{x}v&=-6&\text{where }v=y^{-1}\text{ and }v'=-y^{-2}y'\\\\
v'-\frac{1}{x}v&=6
\end{align*}\]
which is a linear equation. Find the integrating factor.

The equation is also separable, so that's another way to solve it.

Answer 2

Oops, forget what I said about separability

Answer 3

ok, so far I already did that and from there I solved for \[\mu(x)=e^{\int\limits1/xdx}\]

Answer 4

which equals to e^(lnx)=x, then I multiplied x to the equation \[\nu \prime-\frac{ 1 }{ x }\nu ^{-1}=6\]

Answer 5

Right, so you have
\[xv'-v=6x\]
The left side is a derivative:
\[\frac{d}{dx}[xv]=6x\]
Integrate both sides
\[xv=\int6x~dx\]
and so on.

Answer 6

which equals to \[xv'-v=6x\], then I took the integral of the right side only since the left side is the resultant of the derivative of the product rule \[xv=\int\limits{6xdx}\]

Answer 7

oops I just noticed that you had replied that, ok well I pretty much solved the problem already, but I just want to make sure that I did it right...or at least to see if others, such as you are getting the same answer.....

Answer 8

so I got \[v=3x+c\] and since y is the reciprocal of v then \[y=\frac{ 1 }{ 3x+c }\] oh and I forgot to mention that the initial condition is y(1)=-6

Answer 9

so therefore, I ended up solving for c, which c=-19/6, therefore,
\[y=\frac{ 1 }{ 3x-\frac{ 19 }{ 6 }}\]

Answer 10

can someone please let me know if you got the same answer!!!!

Answer 11

\[v \prime-\frac{ 1 }{ x }v=6\]
\[I.F.=e ^{\int\limits \frac{ -1 }{ x }dx}=e ^{-\ln x}=e ^{\ln x ^{-1}}=x ^{-1}=\frac{ 1 }{ x }\]
\[v*\frac{ 1 }{ x }=\int\limits 6*\frac{ 1 }{ x }dx+c\]
\[y ^{-1}*\frac{ 1 }{ x }=6\ln x+c\]