What is the solution for (3(y+1)^2)/(x^4)dx-((2x)(y+1))/(x^4)dy=0 ?

The solution for (3(y+1)^2)/(x^4)dx-((2x)(y+1))/(x^4)dy=0 is y=c_{1}x^{3/2}-1

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

\frac{3 ( y + 1 ) ^{2}}{x ^{4}} d x - \frac{( 2 x ) ( y + 1 )}{x ^{4}} d y = 0

y = c_{1} x^{\frac{3}{2}} - 1

Solution steps

\frac{3 ( y + 1 ) ^{2}}{x ^{4}} d x - \frac{( 2 x ) ( y + 1 )}{x ^{4}} d y = 0

Solve separable ODE:

y = c_{1} x^{\frac{3}{2}} - 1

y = c_{1} x^{\frac{3}{2}} - 1

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What is the solution for (3(y+1)^2)/(x^4)dx-((2x)(y+1))/(x^4)dy=0 ?
The solution for (3(y+1)^2)/(x^4)dx-((2x)(y+1))/(x^4)dy=0 is y=c_{1}x^{3/2}-1