Nernoulli equation y'+6xy = 3xy^3, y(0)=5
what is the answer for this one??

Question

Nernoulli equation y'+6xy = 3xy^3, y(0)=5 
what is the answer for this one??

inkyvoyd · Answer

Use wolfram alpha if you want to find the answer. Use wolfram alpha show steps if you want the process. LOL.

inkyvoyd · Answer

http://www.wolframalpha.com/input/?i=y%27%2B6xy+%3D+3xy^3%2C+y%280%29%3D5+

anonymous · Answer

Make a substitution of u=y^(1-3) which makes it linear See http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/ode/first/bernoulli/bernoulli.html

anonymous · Answer

I don't understand

anonymous · Answer

$$y'+6xy=3xy^3$$
$$v=y^{1-3}=y^{-2}\
v'=-2y^{-3}y'\
y'=-\frac{1}{2}y^3v'$$
$$-\frac{1}{2}y^3v'+6xy=3xy^3\
-\frac{1}{2}v'+6xy^{-2}=3x\
-\frac{1}{2}v'+6xv=3x\
v'-12xv=-6x$$
Now you've got yourself a linear ODE. Find the integrating factor and solve.