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 (6x+1)y2dy/dx+3x2+2y3=0, y(0)=1
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un4t5o4v
Answered 2020-11-17 Author has 105 answers

(6x+1)y2dy/dx+3x2+2y3=0
(6x+1)y2dy+(3x2+2y3)dx=0
(6xy2+1y2)dy+(3x2+2y3)dx=0
Comparing with mdx+ndy=0
n=(6xy2+1y2)
m=(3x2+2y3)
(del m)/(del y)=6y2
(del n)/(del x)=6y2
 mdx+ terms of n not involving x dy= C(x)
(3x2+2y3)dx+y2dy=C(x)
x3+2xy3+y3/3=C(x)
Now applying initial conditions
Substituting the value for C
y(0)=1
x3+2xy3+y3/3=C(x)
C=1/3
x3+2xy3+y3/3=1/3

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