`x^2 + 5y dy/dx = 0` Find the general solution of the differential equation

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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`


Multiply both side by 2/5, we get:


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



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