# Solve differential equation (6x+1)y^2 dy/dx+3x^2+2y^3=0, y(0)=1

Question
Solve differential equation $$(6x+1)y^2 dy/dx+3x^2+2y^3=0$$, y(0)=1

2020-11-17
$$(6x+1)y^2 dy/dx+3x^2+2y^3=0$$
$$(6x+1)y^2dy+(3x^2+2y^3)dx=0$$
$$(6xy^2+1y^2)dy+(3x^2+2y^3)dx=0$$
Comparing with $$mdx+ndy=0$$
$$n= (6xy^2+1y^2)$$
$$m= (3x^2+2y^3)$$
$$(del m)/(del y)=6y^2$$
$$(del n)/(del x)=6y^2$$
int mdx+ int terms of n not involving x dy= C(x)
$$int (3x^2+2y^3)dx+ int y^2 dy= C(x)$$
$$x^3+2xy^3+y^3/3= C(x)$$
Now applying initial conditions
Substituting the value for C
y(0)=1
$$x^3+2xy^3+y^3/3= C(x)$$
$$C=1/3$$
$$x^3+2xy^3+y^3/3=1/3$$

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