Recall that an **ordinary differential equation (ODE)** has differential equation for a function with single variable. A** first order ODE** follows `(dy)/(dx) = f(x,y)` .

In the given problem: `x^2+5y(dy)/(dx)=0` , we apply **variable separable** differential equation in a form of `f(y) dy = f(x) dx` .

Move the `x^2`...

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Recall that an **ordinary differential equation (ODE)** has differential equation for a function with single variable. A** first order ODE** follows `(dy)/(dx) = f(x,y)` .

In the given problem: `x^2+5y(dy)/(dx)=0` , we apply **variable separable** differential equation in a form of `f(y) dy = f(x) dx` .

Move the `x^2` to the other side: `5y(dy)/(dx)=-x^2`

Transfer the `(dx)` to the other side by cross-multiplication: `5y dy=-x^2 dx`

Apply **direct integration**: `int5y dy=int-x^2 dx`

Apply the basic integration property: `int c*f(x) dx= c int f(x) dx` .

`5int y dy=(-1) intx^2 dx`

Apply **Power Rule of integration**: `int x^ndx= x^(n+1)/(n+1)` .

`5*y^(1+1)/(1+1)=(-1) * x^(2+1)/(2+1)+C`

`(5y^2)/2=-x^3/3+C`

Multiply both side by 2/5, we get:

`(2/5)(5y^2)/2=(2/5)(-x^3/3+C)`

Note: `(2/5)*C = C` since `C` is an arbitrary constant.

`y^2=(-2x^3)/15+C`

`y=+-sqrt(-(2x^3)/15+C)`