Question

What is a solution to the differential equation `(1+y)dy/dx-4x=0`?

Answer 1

The solution is `y=+-sqrt(2x^2+1+C_1)-1`

Explanation:

`(1+y)dy/dx-4x=0`

This is a first order ordinary differential equation of the form

`N(y)dy=M(x)dx`

Therefore,

`(1+y)dy/dx=4x`

`(1+y)dy=4xdx`

Integrating both sides

`int(1+y)dy=int4xdx`

`y+y^2/2=4x^2/2+C`

`y^2/2+y=2x^2+C`

`y^2+2y=2x^2+C_1`

`y^2+2y+1=2x^2+1+C_1`

`(y+1)^2=2x^2+1+C_1`

`(y+1)=+-sqrt(2x^2+1+C_1)`

`y=+-sqrt(2x^2+1+C_1)-1`

Answer 2

See below.

Explanation:

`1/2(d/(dx))(1+y)^2-2(d/(dx))x^2=(d/dx)(1/2(1+y)^2-2x^2)=0`

then

`1/2(1+y)^2-2x^2=C` or

`(1+y)^2=4x^2+C_1` or

`1+y=pmsqrt(4x^2+C_1)` or

`y = -1pmsqrt(4x^2+C_1)`