Question

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

First order differential equations
ANSWERED
asked 2020-11-16
Solve differential equation \((6x+1)y^2 dy/dx+3x^2+2y^3=0\), y(0)=1

Answers (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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