What is a general solution to the differential equation $\frac{d y}{d x} = \frac{x^{3}}{y^{2}}$ $dy/dx=x^3/y^2$ ?

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Answer 1 · 2016-09-16T03:33:28

$y = \sqrt[3]{\frac{3 x^{4}}{4} + C}$ $y=root3((3x^4)/4+C)$

Treat $\frac{d y}{d x}$ $dy/dx$ like a fraction to move the $y$ $y$ terms to one side of the equation and the $x$ $x$ terms to the other:

$\frac{d y}{d x} = \frac{x^{3}}{y^{2}} \Rightarrow y^{2} d y = x^{3} d x$ $dy/dx=x^3/y^2" "=>" "y^2dy=x^3dx$

Integrate both sides:

$\Rightarrow \int y^{2} d y = \int x^{3} d x$ $=>" "inty^2dy=intx^3dx$

Using the typical integration power rule:

$\Rightarrow \frac{y^{3}}{3} = \frac{x^{4}}{4} + C$ $=>" "y^3/3=x^4/4+C$

Solving for $y$ $y$ :

$\Rightarrow y^{3} = \frac{3 x^{4}}{4} + C$ $=>" "y^3=(3x^4)/4+C$

$\Rightarrow y = \sqrt[3]{\frac{3 x^{4}}{4} + C}$ $=>" "y=root3((3x^4)/4+C)$

What is a general solution to the differential equation dy/dx=x^3/y^2dydx=x3y2dy/dx=x^3/y^2?