Find the general solution of the differential equation xy'+2y=0

Printable View

- Mar 26th 2007, 05:24 PMvc15ao4Calculus Homework Help Please
Find the general solution of the differential equation xy'+2y=0

- Mar 26th 2007, 05:28 PMJhevon
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 - Mar 26th 2007, 05:28 PMThePerfectHacker
Say x>0, otherwise it gets me angry :mad:

Then,

y'+(2/x)y=0

y'=(-2/x)y

Well y=0 is a solution. Assume y!=0.*

y'/y=(-2/x)

INT (y'/y) dx = INT (-2/x) dx

ln |y| = -2*ln (x)+C

y= exp(-2*ln(x)+C)=C/x^2 where C>0

*)The case y=0 at a point does not need to be considered. - Mar 26th 2007, 05:31 PMvc15ao4i don't know
can somebody doublecheck their answer because my professor told me that y=x^-2 was not the correct answer

- Mar 26th 2007, 05:32 PMtopsquark
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 - Mar 26th 2007, 05:34 PMThePerfectHacker
- Mar 26th 2007, 05:37 PMJhevon
- Mar 26th 2007, 05:41 PMvc15ao4thanks
plugging in the first derivitive did help. what about this problem

Evaluate the integral: int (t^2+1/t+1)dt - Mar 26th 2007, 05:48 PMThePerfectHacker
- Mar 26th 2007, 05:51 PMJhevon
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 - Mar 26th 2007, 05:59 PMvc15ao4???
how come your answer and the perfecthacker's answer were so different

- Mar 26th 2007, 06:00 PMJhevon
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 - Mar 26th 2007, 06:08 PMJhevon
- Mar 26th 2007, 06:09 PMvc15ao4its ok
no it should be fine but thanks anyway