What is the solution for 7x^2+(dy)/(dx)(x^3+6)=0 ?

The solution for 7x^2+(dy)/(dx)(x^3+6)=0 is y=-7/3 ln(x^3+6)+c_{1}

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7x^2+(dy)/(dx)(x^3+6)=0

7 x^{2} + \frac{d y}{d x} (x^{3} + 6) = 0

y = - \frac{7}{3} ln (x^{3} + 6) + c_{1}

Solution steps

7 x^{2} + \frac{d y}{d x} (x^{3} + 6) = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

7 x^{2} + y^{^{'}} (x) (x^{3} + 6) = 0

Isolate

y^{'} (x) : y^{'} (x) = - \frac{7 x ^{2}}{x ^{3} + 6}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - \frac{7 x ^{2}}{x ^{3} + 6} d x

\int - \frac{7 x ^{2}}{x ^{3} + 6} d x = - \frac{7}{3} ln (x^{3} + 6) + c_{1}

y = - \frac{7}{3} ln (x^{3} + 6) + c_{1}

\int \frac{1}{x - x + 6} d x

\frac{d}{d x} (4 (1 - sin (x)))

\frac{\partial}{\partial x} (cos^{2} (x) + cos^{2} (y))

\int_{1}^{2} 6 x^{4} ln (x) d x

f (x) = x^{3} ln (x)

What is the solution for 7x^2+(dy)/(dx)(x^3+6)=0 ?
The solution for 7x^2+(dy)/(dx)(x^3+6)=0 is y=-7/3 ln(x^3+6)+c_{1}