Find the General Solution to Differential Equation X2y'=y2+3xy+X2

[franky2727]

starting with question of find the general sollution of the differential equation

X²y'=y²+3xy+X² would an acceptable answer be y=-x(ln|x|+c+1) i would show all my working but my camera isn't working so i'll save you must the trouble and just skip to a part that i know is correct where dx/x=(v+1)^-2dv

where v=y/x. thanks in advance i think its right just want a second oppinion before plowing it into an exam

 

[Defennder]

I don't understand your notation. Is 'X' different from 'x'?

 

[franky2727]

no just a bit daft with the caps lock button

 

[HallsofIvy]

starting with question of find the general sollution of the differential equation

X²y'=y²+3xy+X² would an acceptable answer be y=-x(ln|x|+c+1) i would show all my working but my camera isn't working so i'll save you must the trouble and just skip to a part that i know is correct where dx/x=(v+1)^-2dv

where v=y/x. thanks in advance i think its right just want a second oppinion before plowing it into an exam

Well, if y= y=-x(ln|x|+c+1), then y'= -(ln|x|+ c+1)- 1. Putting those into the differential equation, x²y'=y²+3xy+x² would give you x²(-ln|x|+ c+ 1)- 1)= (ln|x|+ c+ 1)²+ 3x(-ln|x|+ c+ 1)+ x². Since the right side is clearly going to involve "(ln|x|)²", I don't see how those are going to be equal.

I suspect you have integrated incorrectly. Yes, this is a homogenous equation and the substitution v= y/x gives
$$\frac{dv}{(v+1)^2}= \frac{dx}{x}$$
Integrating both sides of that gives
$$-\frac{1}{v+1}= ln|x|+ C$$
so
$$v+1= -\frac{1}{ln|x|+ C}$$
$$\frac{y}{x}= -1-\frac{1}{ln|x|+ C}$$
$$y= -x-\frac{x}{ln|x|+ C}$$

 

[Defennder]

I worked it out and did not get the same answer as you did. ln(x)+c is supposed to be the denominator of some fraction in y.

 

[franky2727]

got it now, was just checking more to see if it would be acceptible to leave logs and things in the answer but made an error in the question