Question

How can I solve this differential equation? `(3+y+2 y^2 sin^2(x))dx+(x+2xy-y sin(2x))dy=0`

`(3+y+2 y^2 sin^2(x))dx+(x+2xy-y sin(2x))dy=0`

Answer 1

See below.

Explanation:

This differential equation is of type

`M(x,y) dx + N(x.y) dy = 0`

with

`d/dy M(x,y) = d/dx N(x,y)`

then `EE V(x,y) = C_0` such that

`d/dxV(x,y) = M(x,y)`
`d/dy V(x,y) = N(x,y)`

and then

`V(x,y) = int\ M(x,y) dx + phi_1(y) = int\ N(x,y) dy + phi_2(x)`

or

`V(x,y) = 3 x + x y + x y^2 - 1/2 y^2 Sin(2 x) + phi_1(y)`
`V(x,y) = x y + x y^2 - 1/2 y^2 Sin(2 x) +phi_2(x)`

and then

`phi_1(y) = 0, phi_2(x) = 3x`

and finally

`V(x,y) = 3 x + x y + x y^2 - 1/2 y^2 Sin(2 x) = C_0`

or the explicit form

`y = (x pm sqrt[2 (C_0 + 3 x) Sin(2 x)])/(sin(2x)-2 x -x (4 C_0 + 11 x) )`

Answer 2

Given: `(3+y+2 y^2 sin^2(x))dx+(x+2xy-y sin(2x))dy=0`

Check for exactness by computing the following partial derivatives:

`(del(3+y+2 y^2 sin^2(x)))/(dely) = 4ysin^2(x)+1`

`(del(x+2xy-y sin(2x)))/(delx)= 2ycos(2x)+2y+1`

Use the identity `cos(2x) = 2sin^2(x)-1`:

`(del(x+2xy-y sin(2x)))/(delx)= 2y(2sin^2(x)-1)+2y+1`

Distribute 2y:

`(del(x+2xy-y sin(2x)))/(delx)= 4ysin^2(x)-2y+2y+1`

`(del(x+2xy-y sin(2x)))/(delx)= 4ysin^2(x)+1`

Because we have observed that `(del(3+y+2 y^2 sin^2(x)))/(dely) = (del(x+2xy-y sin(2x)))/(delx)`, we know that the differential equation is exact. The solution method for an exact differential equation is well known.

We know that the solution is of the `f(x,y) = C` and we know that:

`(del(f(x,y)))/(delx) = 3+y+2 y^2 sin^2(x)" [1]"`

and

`(del(f(x,y)))/(dely) = x+2xy-y sin(2x)" [2]"`

We can choose to integrate either equation; I shall choose to integrate equation [2]:

`f(x,y) = int x+2xy-y sin(2x)dy`

`f(x,y) = xy+xy^2-1/2y^2sin(2x)+ phi(x)" [3]"`

where `phi(x)` is a function of x that becomes 0, when `f(x,y)` is differentiated with respect to y.

To find `phi(x)`, we must differentiate equation [3] with respect to x:

`(del(f(x,y)))/(delx) = y+y^2+y^2cos(2x)+ phi'(x)`

Use the identity `cos(2x) = 2sin^2(x) -1`:

`(del(f(x,y)))/(delx) = y+y^2+y^2(2sin^2(x) -1)+ phi'(x)`

`(del(f(x,y)))/(delx) = y+2y^2sin^2(x)+ phi'(x)`

We know that right side of the above equation must equal the right side of equation [1]:

`y+2y^2sin^2(x)+ phi'(x)= 3+y+2 y^2 sin^2(x)`

Solve for `phi'(x)`:

`phi'(x) = 3`

`phi(x) = 3x`

Substitute this into equation [3]:

`f(x,y) = xy+xy^2-1/2y^2sin(2x)+ 3x`

Make the equation fit the known form by setting it equal to an arbitrary constant C:

`f(x,y) = xy+xy^2-1/2y^2sin(2x)+ 3x= C`