Question

find the particular solution of (x^2 + y^2)y' + 2xy = 0 satisfying y(1)=1

Answer 1

\[\left( x ^{2} + y^{2} \right) y \prime +2xy = 0\]

Answer 2

it is a homogenous differential equation
\[dy/dx=\-\2xy/x \^\{\2\}\+y \^\{\2\}\]

Answer 3

subs y=vx

Answer 4

the answer is
y=lnx+(1/3)ln((y/x)^3 +3(y/x))+C

y(1)=1
1=ln1+(1/3)ln(1^3 +3)+C
1=0+(1/3)ln4 +C
C=1-(1/3)ln4=0.538
y=lnx+(1/3)ln((y/x)^3 +3(y/x))+0.538