Thread: Calculus Homework Help Please

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

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

Originally Posted by vc15ao4

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

Say x>0, otherwise it gets me angry

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.

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

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

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

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

plugging in the first derivitive did help. what about this problem

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

Originally Posted by vc15ao4

plugging in the first derivitive did help. what about this problem

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)

Originally Posted by vc15ao4

plugging in the first derivitive did help. what about this problem

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

how come your answer and the perfecthacker's answer were so different

Originally Posted by vc15ao4

how come your answer and the perfecthacker's answer were so different

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

Originally Posted by vc15ao4

how come your answer and the perfecthacker's answer were so different

i can come up with yet another solution if you want.

was there a particular form you want the solution in?

no it should be fine but thanks anyway