# Answer to Question #20864 in Differential Equations for behzad

Question #20864

y^2dx=(x^3-xy)dy

Expert's answer

y^2dx=(x^3-xy)dx

xy=u

u'=x'y+x

y^2x'(y)=(x^3-xy)dx

x'(y)=u'/y+x

yu'(y)-xy=x^3-xy

yu'(y)=u^3(y)/y^3

y^4u'(y)=u^3(y)

y^-4dy=u^-3du

y^-3/3=u^-2/2+C

u=Sqrt(1.5y^3+C)

x=u/y

x(y)=Sqrt(1.5y^3+C)/y

xy=u

u'=x'y+x

y^2x'(y)=(x^3-xy)dx

x'(y)=u'/y+x

yu'(y)-xy=x^3-xy

yu'(y)=u^3(y)/y^3

y^4u'(y)=u^3(y)

y^-4dy=u^-3du

y^-3/3=u^-2/2+C

u=Sqrt(1.5y^3+C)

x=u/y

x(y)=Sqrt(1.5y^3+C)/y

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