Question

Solve y'=(y^3+2xy^2+x^2*y+x^3)/(x(y+x)^2) implicitly.

Answer 1

y=ux
u=y/x
y'=u+u'x
u'x+u=((ux)^3+2x(ux)^2+x^2(ux)+x^3))/(x(ux+x)^2)
u'x+u=((ux)^2(ux+2x)+x^2(ux+x))/(x(ux+x)^2)
How do I simplify this? The answer in the book is (y+x)^3=3x^3(ln abs(x)+C). How do I get to the answer from there?

Answer 2

@amistre64

Answer 3

x cancels out everytime you substitute `y=ux` in a homogeneous equation

Answer 4

u'x+u = ((ux)^3+2x(ux)^2+x^2(ux)+x^3)) / (x(ux+x)^2)

first cancel x

Answer 5

What do I get if I cancel x?

Answer 6

you will get a separable equation

Answer 7

thats the only idea behind y=ux substitution

Answer 8

Let me factor out first. Please don't leave.

Answer 9

okie

Answer 10

I don't see how I can cancel out the x.
From u'x+u=((ux)^3+2x(ux)^2+x^2(ux)+x^3)/(x(ux+x)^2)

u'x+u=((ux)^2(ux+2x)+x^2(ux+x))/(x(ux+x)^2)
u'x+u=(x(ux)^2(u+2)+x^2(ux+x))/(x(ux+x)^2)

Answer 11

\[\large \begin{align}\\
u'x+u = \dfrac{(ux)^3+2x(ux)^2+x^2*(ux)+x^3}{x(ux+x)^2}

\end{align}\]

Answer 12

Notice that each term in the numerator of right hand side has a factor x^3

Answer 13

\[\large \begin{align}\\
u'x+u &= \dfrac{(ux)^3+2x(ux)^2+x^2*(ux)+x^3}{x(ux+x)^2} \\~\\
&= \dfrac{x^3[u^3+2u^2+u+1]}{x^3(u+1)^2} \\~\\

\end{align}\]

Answer 14

But how do I simplify (u^3+2u^2+u+1)/(u+1)^2 after I cancel out the x^3?

Answer 15

not sure yet, but thats just an algebra problem

Answer 16

maybe send that u to right side first

Answer 17

\[\large \begin{align}\\
u'x+u &=\dfrac{u^3+2u^2+u+1}{(u+1)^2} \\~\\
u'x &=\dfrac{u^3+2u^2+u+1}{(u+1)^2} - u \\~\\

\end{align}\]

Answer 18

get common denominator and simplify

Answer 19

Got it! Thanks a lot!