• Mar 26th 2007, 04:24 PM
vc15ao4
Find the general solution of the differential equation xy'+2y=0
• Mar 26th 2007, 04:28 PM
Jhevon
Quote:

Originally Posted by vc15ao4
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
• Mar 26th 2007, 04:28 PM
ThePerfectHacker
Quote:

Originally Posted by vc15ao4
Find the general solution of the differential equation xy'+2y=0

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, 04:31 PM
vc15ao4
i 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, 04:32 PM
topsquark
Quote:

Originally Posted by vc15ao4
Find the general solution of the differential equation xy'+2y=0

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, 04:34 PM
ThePerfectHacker
Quote:

Originally Posted by vc15ao4
can somebody doublecheck their answer because my professor told me that y=x^-2 was not the correct answer

That is because it is a solution.
It is not general solution.

Meaning, all solution.s
• Mar 26th 2007, 04:37 PM
Jhevon
Quote:

Originally Posted by vc15ao4
can somebody doublecheck their answer because my professor told me that y=x^-2 was not the correct answer

the answer is a constant times x^-2 or C/x^2. you can check this answer yourself by finding the first derivative and plugging it in to the function
• Mar 26th 2007, 04:41 PM
vc15ao4
thanks

Evaluate the integral: int (t^2+1/t+1)dt
• Mar 26th 2007, 04:48 PM
ThePerfectHacker
Quote:

Originally Posted by vc15ao4

Evaluate the integral: int (t^2+1/t+1)dt

(t^2+1)/(t+1)=(t^2+2t-2t/(t+1)=(t^2+2t)/(t+1)-2*(t)/(t+1)=t-2*(t/(t+1))

But,
t/(t+1)=(t+1-1/(t+1))=1-1/(t+1)

Thus, the partial fractions are,

t+1-1/(t+1)
• Mar 26th 2007, 04:51 PM
Jhevon
Quote:

Originally Posted by vc15ao4

Evaluate the integral: int (t^2+1/t+1)dt

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, 04:59 PM
vc15ao4
???
• Mar 26th 2007, 05:00 PM
Jhevon
Quote:

Originally Posted by vc15ao4

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, 05:08 PM
Jhevon
Quote:

Originally Posted by vc15ao4