Differential Equations help!

[Marsxtc]

I'm trying to find a general equation, y= (fx) of dy/dx = x(2y+1)
I got sqrt(2y+1) = Ce^(x^2/2) Where C = +/- e^c
I'm not sure if I got this answer correct.
Before I exponentiated both sides I had 1/2 ln(abs(2y+1)) = x^2/2+C.
With my initial conditions of f(2) = 0. I got C = -2.

 

[Deleted member 4993]

I'm trying to find a general equation, y= (fx) of dy/dx = x(2y+1)
I got sqrt(2y+1) = Ce^(x^2/2) Where C = +/- e^c
I'm not sure if I got this answer correct.
Before I exponentiated both sides I had 1/2 ln(abs(2y+1)) = x^2/2+C.
With my initial conditions of f(2) = 0. I got C = -2.

Differentiate implicitly:

(2y+1)^1/2 = Ce^(x²/2) and see if it satisfies the original equation or not.

If not - you made a mistake somewhere.

If yes - you are most probably correct.

 

[Marsxtc]

Wow , I tried implicitly differentiating the right side, and it looks nothing like the original. I think i'm done for haha...
Trying to do it again
Dy/dx = x(2y+1)
dy = x(2y+1)dx
dy/2y+1 = xdx
Integrate both sides
1/2lnabs(2y+1) = x^2/2+C
and then exponentiating both sides
sqrt(2y+1) = e^(x^2/2 +C)
sqrt(2y+1) = C e^(x^2/2)
I dont know what i'm doing wrong...

 

[Deleted member 4993]

Wow , I tried implicitly differentiating the right side, and it looks nothing like the original. I think i'm done for haha...
Trying to do it again
Dy/dx = x(2y+1)
dy = x(2y+1)dx
dy/2y+1 = xdx
Integrate both sides
1/2lnabs(2y+1) = x^2/2+C
and then exponentiating both sides
sqrt(2y+1) = e^(x^2/2 +C)
sqrt(2y+1) = C e^(x^2/2)
I dont know what i'm doing wrong...

sqrt(2y+1) = C e^(x^2/2)

2y + 1 = C²e^(x²)

y = K*(e^x²) - ½ .... where K = C²/2

y' = K*2*x*(e^x²) = 2 * C²/2 * x * (e^x²) = C² * x * (e^x²) - x + x = x * [C² * x * (e^x²) - 1] + x = x * 2y + x = x * (2y + 1)

You give up too easy....