Question

\[y^2+2y=\ln |x| +C\] Given that y(1)=-2

Solve the equation in terms of y

Answer 1

I'm trying to solve the differential equation
\[ (1+y)* \frac{dy}{dx} =\frac{1}{2x}\]

where \( y(1)=-2\)

I have separated the equation and taken the derivative on either side but hit a roadblock when I attempt to formalize my response in an explicit form.

Answer 2

You dont need explicit form to find constant C
\[(-2)^2 +2(-2) = \ln(1) + C\]
C = 0
Finally to solve for y, i would use completing the square since y is a quadratic
\[(y+1)^2 = \ln x +1\]
\[y = \pm \sqrt{\ln x + 1} - 1\]

Answer 3

y^2+2y = ln(|x|) + C
y^2+2y+1 = ln(|x|) + C+1 ... add 1 to both sides to complete the square
(y+1)^2 = ln(|x|) + C+1
y+1 = +-sqrt( ln(|x|) + C+1 )
y = -1+-sqrt( ln(|x|) + C+1 )

Answer 4

ahhhhhh completing the square =_=
Should have thought of that
Also only negative squareroot would work with the initial condition

Answer 5

\[-2 = - \sqrt{\ln (1) + 1} - 1\]

Answer 6

Yeah, looks the top part of parabola must be discarded

Answer 7

Yep, y(x)=-1-sqrt(1+log|x|), with y(1)=-2
and substitute into original,
(1+y(x))dy/dx=1/(2x)....yay!