Question #fb874

1 Answer
HCH
Jul 11, 2017

#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#.

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