What is the general answer to this differential equation?2x(1+y^2)dx-y(1+2x^2)dy?

Question

I need all the solution :)

am really sorry ive forgotten to put the equation at the end thank you for your recommendations so ....2x(1+y^2)dx-y(1+2x^2)dy=0

Answer 1

check if it is exact
@=delta
(2x+2xy^2)dx
+(-y+2yx^2)dy
@M/@y=2x+4xy
@N/@x=-y+4xy
therefore not exact

use separable
2xdx/(1+2x^2)-ydy/(1+y^2)=0
u=(1+2x^2) u=(1+y^2)
du=4x du=2y

1/2 ln[(1+2x^2)]- 1/2 ln [(1+y^2)]=ln c
property of ln we will get
ln[(1+2x^2)^1/2]- ln [(1+y^2)^1/2]=ln c
property of ln we will get
ln[(1+2x^2)^1/2]/[(1+y^2)^1/2]=ln c
cancel the ln- that is also a property

[(1+2x^2)/(1+y^2)]^1/2=c
this is the general solution

somewhat in other differential equation u must know if u use what technique, if it is separable, homogeneous, exact, linear,
determination of integral factors, etc. So u know what it is. U also must have a background of integral and differential calculus.

i answered this question cause i see the proper way of solving it by other people.
i realized the way then i arrived to this answer

Answer 2

I assume it is equal to zero:

2x(1 + y^2) dx - y(1 + 2x^2)dy = 0

y(1 + 2x^2)dy = 2x(1 + y^2) dx

this equation is SEPERABLE (all x and dx on one side all y and dy on the other)

ydy/(1 + y^2) = 2xdx/(1 + 2x^2)

Integrate both sides:

∫ydy/(1 + y^2) = ∫2xdx/(1 + 2x^2)

For the LHS let u = 1 + y^2 you get ydy = du/2
For the RHS let w = 1 + 2x^2 you get 2xdx = dw/2

and the equation becomes:

1/2∫ du/u= 1/2∫ dw/w

These both integrate as natural log

ln(u) = ln(w)

so u = Aw where A is the arbitrary constant

1 + y^2 = A(1 + 2x^2)

y = √(A + 2Ax^2 -1)

so suppose dy/dy = 4/3 when x = 1

you find A = 2/3

and you get: y = √(4/3x^2 - 1/3) as the solution

CHECK: if you compute dy/dx of √(4/3x^2 - 1/3) and solve for dy/dx in 2x(1 + y^2) dx - y(1 + 2x^2)dy = 0 when y = √(4/3x^2 - 1/3) you'll get equality and you know the solution MUST be correct.

Answer 3

This equation is called a separable first-order equation because it can be written in the form f(y) dy=g(x) dx:

2x(1+y^2)dx-y(1+2x^2)dy=0

y(1+2x^2)dy=2x(1+y^2)dx

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

Now integrate the left side with respect to y and the right side with respect to x (u=1+y^2 on the left; u=1+2x^2 on the right)

1/2 ln(1+y^2)=1/2 ln (1+2x^2)+C

Multiply through by 2

ln(1+y^2)=ln(1+2x^2)+C

Exponentiate:

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

Solve for y:

y^2=C(1+2x^2)-1
y= \pm sqrt{C(1+2x^2)-1}

PS Note: you should always check linearity first. This equation is not linear in y because of the y^2 term and it is not linear in x because of the x^2 term. Do not solve this equation as though it were a linear equation--it would be a "fatal" error in a testing situation and you almost certainly wouldn't get any partial credit! :)

Answer 4

You know, for something to be an equation, an equal sign(=) is vital. You seem to have skipped grade school.
However,
Whatever it may equal, divide throughout by -y(1+ 2x^2)dx to make the dy/dx term have a coefficient of unity. Then rearrange in the form,
dy/dx + Py = Q
where P and Q are functions of x,
then get
e^ integral P dx
Multiply throughout by this and you'll notice an exact product
d[(e^ integral Pdx) . y] / dx = (e^ integral Pdx).Q
Seperate variables, integrate, divide by coefficient of y and voila.

Answer 5

2x(1+y^2)dx-y(1+2x^2)dy
= (2x+2xy^2)dx - (y+2x^2)dy
= 2dx^2+(2d x^2 y^2) - dy^2 -(2d x^2 y^2)
= 2dx^2 - dy^2

Answer 6

yeah, the hole ellipse makes it sort of hard to answer.

Answer 7

i need all the question.

Answer 8

Where is question? I see this (.......)