Question

Question #fb874

Answer 1

`x^4=-4(y^3/3+y^2+y)+C_2`

Explanation:

`x^3+(y+1)^2dy/dx=0`

Rearrange,
`x^3=-(y+1)^2dy/dx`

Separate `dy` and `dx`,
`x^3dx=-(y+1)^2dy`

Integrate both sides,
`intx^3dx=-int(y^2+2y+1) dy`
note that we can take out coefficient -1 on the RHS

`x^4/4=-(y^3/3+y^2+y)+C_1`
note that I only put C (constant) on one side as C from LHS - C from RHS (or vice versa) is another constant;
the same rule applies to when I didn't include it in a bracket

`x^4=-4(y^3/3+y^2+y)+C_2`
note that I put a subscript for C, `C_1 != C_2`

I assume this is simpler enough (if not simpler than the differential equation), you could proceed on if you want a pure `x` in terms of `y`.