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Title: How to solve the differential equation 2x(x+y)y'=3y^2+4xy for y(1)=1? Post by: nvaldivi2 on Nov 13, 2012 I use substitution y=vx , but I get stucked at (2+2v)/(v^2+2v) dv = 1/x dx.
Thanks for the help!!! Title: How to solve the differential equation 2x(x+y)y'=3y^2+4xy for y(1)=1? Post by: LessieP on Nov 13, 2012 U R right.
But you get sucked because you did not notice that (2+2v)/(v^2+2v) = 2/v after simplication. Simple algebra!!!! Title: How to solve the differential equation 2x(x+y)y'=3y^2+4xy for y(1)=1? Post by: buggingout16 on Nov 14, 2012 its a homogenous D.E.
it can be written as y'=(3y^2+4xy)/(2x^2+2xy). Now on RHS divide numerator and denominator by highest power of x present in the equation i.e. x^2. We get y'=[3(y/x)^2+4(y/x)]/[2+2(y/x). Now substitute y/x =t on differtiating both sides we get. y'=t+xt' now overall eq becomes: t+xt' = (3t^2+4t)/(2+2t) by variable seperable: (t^2+2t)/(2+2t) dt = (dx)/x integrate both sides to get anwser |