I have the following equation(xy2+x)dx+(yx2+y)dy=0and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method.Let M=xy2+x and N=yx2+yMy=2xy and Nx=2xy∫M.dx⇒∫xy2+x=x2y2+x22+g(y)Partial of (x2y2+x22+g(y))⇒xy2+g(y)′g(y)′=yg(y)=y22the general solution then isC=x2y22+x22+y22Is this solution the same I would get if I had taken the Separate Equations route?

Taking the other route you get y′y(x2+1)=−xy2−x.Denote z=y2, hence z′=2yy′. So you get z′+2xx2+1z=−xx2+1. This gives you z=k(x)x2+1nSKSubstituting, you have k′(x)=−x, which implies k(x)=−x22+C. Hence z=−x22(x2+1)+Cx2+1. Finally, y2=zFrom here you get 2y2(x2+1)+x2=C, which is the answer you got.

"I have the following equation(xy2+x)dx+(yx2+y)dy=0and I am told..." was answered by our experts

veksetz

Answered question

2022-01-18

I have the following equation

(x y^{2} + x) d x + (y x^{2} + y) d y = 0

and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method.
Let

M = x y^{2} + x and N = y x^{2} + y

M y = 2 x y and N x = 2 x y

\int M . d x \Rightarrow \int x y^{2} + x = x^{2} y^{2} + \frac{x^{2}}{2} + g (y)

Partial of

(x^{2} y^{2} + \frac{x^{2}}{2} + g (y)) \Rightarrow x y^{2} + {g (y)}^{'}

{g (y)}^{'} = y

g (y) = \frac{y^{2}}{2}

the general solution then is

C = x^{2} \frac{y^{2}}{2} + \frac{x^{2}}{2} + \frac{y^{2}}{2}

Is this solution the same I would get if I had taken the Separate Equations route?

Answer & Explanation

Annie Levasseur

Beginner2022-01-19Added 30 answers

Basically, yes. You would write

\int \frac{y d y}{y^{2} + 1} = \int - \frac{x d x}{x^{2} + 1}

\frac{1}{2} \ln (y^{2} + 1) = - \frac{1}{2} \ln (x^{2} + 1) + c

Multiply both sides by 2 and exponentiate:

y^{2} + 1 = \frac{e^{2 c}}{x^{2} + 1}

This is equivalent to your solution, with

C = \frac{e^{2 c} - 1}{2}

Chanell Sanborn

Beginner2022-01-20Added 41 answers

Taking the other route you get

y^{'} y (x^{2} + 1) = - x y^{2} - x

.
Denote

z = y^{2}

, hence

z^{'} = 2 y y^{'}

. So you get

z^{'} + \frac{2 x}{x^{2} + 1} z = - \frac{x}{x^{2} + 1}

. This gives you

z = \frac{k (x)}{x^{2} + 1}

nSKSubstituting, you have

k^{'} (x) = - x

, which implies

k (x) = - \frac{x^{2}}{2} + C

. Hence

z = - \frac{x^{2}}{2 (x^{2} + 1)} + \frac{C}{x^{2} + 1}

. Finally,

y^{2} = z

From here you get

2 y^{2} (x^{2} + 1) + x^{2} = C

, which is the answer you got.

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