Question

A function y(t) satisfies the differential equation dy/dx=y^4-6y^3+5y^2...What are the constant solutions of the equation?

Answer 1

I recommend you to separate this DE, and then use partial fraction decompositions.

Answer 2

Yeah I know to use partial fractions...My problem is how I should separate the two...Would it be like dy/(-y^4+6y^3-5y^2)=dx

Answer 3

Then find the integral of that?

Answer 4

\[\frac{ dy }{ y^4-6x^2+5y^2 }=\frac{ dy }{ y^2(y-5)(y-6) }=dx\]

Answer 5

I made a slight erroro while typing there, it should be (y-5)(y-1) of course.

Answer 6

\[

\frac{ dy }{ y^4-6x^2+5y^2 }=\frac{ dy }{ y^2(y-5)(y-1) }=dx
\]

Answer 7

How would I take the integral of that though?

Answer 8

you need to apply partial fraction decomposition, that will give you a linear integral expression which can again be simplified.

Answer 9

dy/y^2+dy/(y-5)+dy/(y-1)=dx

Answer 10

careful, you can't split up the integral like that, partial fraction decomposition is a bit more complicated than that.

Answer 11

The partial fraction decomposition process for this equation is a bit lengthy, however there are good tutorials all around that give a good introduction to it.

Unfortunately I have to go AFK for a little bit, otherwise I could navigate you through the process, so I will only write down the basic steps:

\[\frac{ 1 }{ y^2(y-5)(y-1) }=\frac{ A }{ y }+\frac{ B }{ y^2 }+\frac{ C }{ (y-5) }+\frac{ D }{ (y-1) }\]

Now solve for A,B,C,D by multiplying the entire equation by the RHS and then use the cover up method (setting one term to zero) or match the coefficients.

You should end up at:

\[\left( \frac{ 1 }{5y^2 }-\frac{ 1 }{ 4(y-1)}+\frac{ 6 }{ 25y }+\frac{ 1 }{ 100(y-5) } \right)dy=dx\]

Answer 12

try working through this steps on your own, I will be back in a few and if you still have troubles then, I might be able to help you.

Answer 13

Thanks :) Lol yeah now I got this completely wild answer! XD\[-\ln(|y-1|)/4 + -\ln(|y-5|)/100 + 6\ln(|y|)/25 - 1/5y = C\]

Answer 14

looks right to me, some terms can be simplified, laws of logarithms, but it's clearly not a pleasant looking solution, and I doubt that it can be expressed explicticallly.

The RHS of your equation isn't quite right there yet: remember!

\[\int\limits dx= \int\limits 1 \cdot dx = x+C\]

Answer 15

Ahhh okay oops yeah I forgot the x :P Thanks :) Haha...Okay so wait, how do we find the constant solutions of this equation?