Discussion:
I am stuck 4xy dx + (x^2+1)dy = 0
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j***@gmail.com
2014-12-08 10:37:15 UTC
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4xy dx + (x^2+1)dy = 0

x^2+1) dy = - 4xy dx
1/2x *(x^2+1) dx = -2ydy

2*ln|x| *(x^2+1)^2 dx = -y^2+c

y^2 + 2*ln|x|*(x^2+1)^2= c





book answer: (x^2+1)^2 y=c

JPD
d***@gmail.com
2017-01-24 01:08:37 UTC
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Post by j***@gmail.com
4xy dx + (x^2+1)dy = 0
x^2+1) dy = - 4xy dx
1/2x *(x^2+1) dx = -2ydy
2*ln|x| *(x^2+1)^2 dx = -y^2+c
y^2 + 2*ln|x|*(x^2+1)^2= c
book answer: (x^2+1)^2 y=c
JPD
You are wrong on third line, it should be
1/y dy/dx= -4x/(x^2+1)
input it into www.mathhandbook.com, click the dsolve button for answer.
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