Question

Differential equation:

Solve the differential equation by regarding y as the independent variable rather than x.

(1+2xy)(dy/dx)= 1+y^2

Answer 1

So i tried to put it into the format of
\[\frac{ dy }{ dx }+P(x)*y=Q(x)\]

Answer 2

(switching basically the x and y components since y is independent) but i keep getting stuck

Answer 3

\[1+2xy=(1+y^2)*\frac{ dx }{dy }\]
\[\frac{ dx }{ dy }-\frac{ 2xy }{ 1+y^2}=\frac{ 1 }{ 1+y^2}\]

Answer 4

so then p(y) = \[\frac{ -2y }{ 1+y^2 }\]
if i integrate that i get
\[-\ln (y^2+1)\]
Then to the e i get
\[-(y^2+1)\]
so then my equation becomes
\[-(y^2+1)x'-2y*x=-1\]

Answer 5

but...i think i did something wrong

Answer 6

the answer should be(solving through the last equation i wrote)
\[=1+y^2)x(y)= \frac{ 1 }{ 2 }[y+(1+y^2)(\tan^{-1} (y)+c)]\]

Answer 7

ignore the stuff infront of x(y)...typo

Answer 8

integrating through i get:
\[-x(y^2+1)=-y+c\]
\[x=\frac{ y+c }{ y^2+1 }\]

..which clearly is wrong

Answer 9

The integrating factor is almost right.

You're correct with
\[\mu(y)=e^{-\ln(y^2+1)}\]
However, this is not equal to \(-(y^2+1)\):
\[e^{-\ln(y^2+1)}=e^{\ln(y^2+1)^{-1}}=\frac{1}{y^2+1}\]

Answer 10

dang it i did that on an earlier problem too...alright ill try that real quick

Answer 11

got it, thank you

Answer 12

Yep, you're welcome