 # Solve differential equation (6x+1)y^2 dy/dx+3x^2+2y^3=0, y(0)=1 generals336 2020-11-16 Answered
Solve differential equation $\left(6x+1\right){y}^{2}dy/dx+3{x}^{2}+2{y}^{3}=0$, y(0)=1
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$\left(6x+1\right){y}^{2}dy/dx+3{x}^{2}+2{y}^{3}=0$
$\left(6x+1\right){y}^{2}dy+\left(3{x}^{2}+2{y}^{3}\right)dx=0$
$\left(6x{y}^{2}+1{y}^{2}\right)dy+\left(3{x}^{2}+2{y}^{3}\right)dx=0$
Comparing with $mdx+ndy=0$
$n=\left(6x{y}^{2}+1{y}^{2}\right)$
$m=\left(3{x}^{2}+2{y}^{3}\right)$

terms of n not involving x dy= C(x)
$\int \left(3{x}^{2}+2{y}^{3}\right)dx+\int {y}^{2}dy=C\left(x\right)$
${x}^{3}+2x{y}^{3}+{y}^{3}/3=C\left(x\right)$
Now applying initial conditions
Substituting the value for C
y(0)=1
${x}^{3}+2x{y}^{3}+{y}^{3}/3=C\left(x\right)$
$C=1/3$
${x}^{3}+2x{y}^{3}+{y}^{3}/3=1/3$

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