Find the general solution of the differential equation xy'+2y=0
If i'm not mistaken, i solved this problem already, anyway...
we proceed using the separation of variables
xy' + 2y = 0
=> y' + (2/x)y = 0
=> y' = (-2/x)y
=> y'/y = -2/x
=> lny = -2ln(x) + C = ln(x)^-2 + C.............integrated both sides
since lny = lnAx^-2
=> y = Ax^-2 is the general solution, where A is a constant
This is separable:
x(dy/dx) + 2y = 0
x dy = -2y dx
dy/y = -2 dx/x
Int[dy/y] = -2*Int[dx/x]
ln(y) = -2*ln(x) + C' <-- C' is a constant
ln(y) + 2*ln(x) = C'
ln(y*x^2) = C'
y*x^2 = C <-- where C = e^{C'}
y = C/x^2 where C is a constant that is not equal to 0.
But note that y = 0 does, in fact, solve xy' + 2y = 0, so we can remove this restriction on C.
-Dan
let u = t + 1
=> du = dt
since u = t + 1
=> t = u - 1
=> t^2 = (u - 1)^2
=> t^2 + 1 = (u - 1)^2 + 1 = u^2 -2u + 2
so our integral becomes:
int{(u^2 -2u + 2)/u}du
= int{u - 2 + 2/u}du
= (1/2)u^2 - 2u + 2ln(u) + C
= (1/2)(t + 1)^2 - 2(t + 1) + 2ln(t+1) + C
= (1/2)(t + 1)( (t + 1) - 4) + 2ln(t + 1) + C
= (1/2)(t + 1)(t - 3) + 2ln(t + 1) + C
that happens sometimes in doing integration, the difference will be absorbed in the arbitrary constants in our respective solutions (or possibly rectified through algebraic manipulations).
my solution is actually the one given by The Integrator--Integrals from Mathematica