Im solving Homogeneous DE.. in the text, im solving this problem... (x^2+y^2)dx
ID: 2986091 • Letter: I
Question
Im solving Homogeneous DE.. in the text, im solving this problem...
(x^2+y^2)dx + (x^2-xy)dy = 0
I understand I have to make a variable subsitution, in this case y =ux...then algebraically solve..
eventually, after simplfying and using long division, I get [-1+2/(1+u)]du +dx/x = 0..
seperable equation, and integrate both sides.. now.. in the text it says after integration you get,
-u+2 ln(1+u) +ln (x) = ln(c)... Where does the ln(c) come from??? I dont get why they have ln(c).. can someone explain this to me??
Explanation / Answer
we assume constant as C in many cases.
in this problem we assume constant as ln(C) for simplification purpose. there is no wrong in assume it as ln(C0 instead C.
(x^2+y^2)dx + (x^2-xy)dy = 0
x du/dx = -(1+u)/(1-u)
==>
du*(1-u)/(1+u) = -x dx
==>
du*[2/(1+u) - 1] = - dx/x
integrate
2*ln(1+u) - u = -ln(x) + ln(C)
=>
ln((1+u)^2) - ln(C) - u = -ln(x)
==>
ln(x(1+u)^2/C) = u
==>
x(1+u)^2/C = e^u
==>
x(1+u)^2 = Ce^u
==>
(x+y)^2 = Cxe^(y/x)