Question

Solve the differential equation:
(1-2x^2-2y)dy/dx=4x^3+4xy

Answer 1

`u=-x^4-2x^2y+y-y^2`

Explanation:

We are given that

`(1-2x^2-2y)(dy)/(dx)=4x^3+4xy`

We have to solve the given differential equation

`(1-2x^2-2y)dy=(4x^3+4xy)dx`

`(1-2x^2-2y)dy-(4x^3+4xy)dx=0`

Compare with
`Mdx+ndy=0`

Then, we get
`M=-(4x^3+4xy),N=(1-2x^2-2y)`

Exact differential equation

`M_y=N_x`

`M_y=-4x`

`N_x=-4x`

`M_y=N_x`

Hence, the differential equation is an exact differential equation.

Solution of exact differential is given by

`u=\int M(x,y)dx+K(y)` where K(y) is a function of y.

`u=\int -(4x^3+4xy) dx+k(y)` y treated as constant

`u=-x^4-2x^2y+k(y)`

`u_y=N`

`-2x^2+k'(y)=1-2x^2-2y`

`K'(y)=1-2y`

`k(y)=y-y^2`

Substitute the value then we get

Then,
`u=-x^4-2x^2y+y-y^2`

Answer 2

(1−2x^2−2y)y′=4x^3+4xy

(1−2x^2−2y)dy=(4x^3+4xy)dx

(4x^3+4xy)dx+(−1+2x^2+2y)dy=0

f(x,y)=C;df(x,y)=∂f/∂x(dx)+∂f/∂y(dy)=0

∂f /∂x=4x^3+4xy⇒f(x,y)=x^4+2x^2y+g(y)

∂f/∂y=−1+2x^2+2y=2x^2+∂g/∂y⇒∂g/∂y=2y−1

g(y)=y^2−y⇒f(x,y)=x^4+2x^2y+y^2−y=(x^2+y)2−y=C