DE Solution: y' -2y/(x+a) = -1 with Integrating Factor (x+a)^-2

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Homework Statement


Solve DE: y' -2 y / (x + a) = -1 where a = constant

Homework Equations


y' + p(x) y = q(x) solving this DE with integrating factor.

The Attempt at a Solution


Use integrating factor (x+a) ^ -2 for above DE,

[ y'(x+a) ^ -2 ]' = - (x+a)^ -2 solving this DE we get

y = C(x+a)^2 + (x+a) this seems to be a solution, C = arbitrary constant

Is this the only solution??
 
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MACH2 said:

Homework Statement


Solve DE: y' -2 y / (x + a) = -1 where a = constant

Homework Equations


y' + p(x) y = q(x) solving this DE with integrating factor.

The Attempt at a Solution


Use integrating factor (x+a) ^ -2 for above DE,

[ y'(x+a) ^ -2 ]' = - (x+a)^ -2 solving this DE we get

y = C(x+a)^2 + (x+a) this seems to be a solution, C = arbitrary constant

Is this the only solution??
Yes, your 1-parameter family of solutions contains every particular solution. In fact, every equation of this form, when solved by multiplying by the integrating factor of the exponential of the integral of p, will yield a 1-parameter family of solutions that contains every particular solution, as long as p and q are continuous on the interval of the solution.
 
MACH2 said:

Homework Statement


Solve DE: y' -2 y / (x + a) = -1 where a = constant

Homework Equations


y' + p(x) y = q(x) solving this DE with integrating factor.

The Attempt at a Solution


Use integrating factor (x+a) ^ -2 for above DE,

[ y'(x+a) ^ -2 ]' = - (x+a)^ -2 solving this DE we get
The above isn't right. The left side is [y * (x + a)-2]', or in another form d/dx[y * (x + a)-2].
MACH2 said:
y = C(x+a)^2 + (x+a) this seems to be a solution, C = arbitrary constant

Is this the only solution??
 
Mark44 said:
The above isn't right. The left side is [y * (x + a)-2]', or in another form d/dx[y * (x + a)-2].

(x+a)^-2 = (x+a) elevated to the -2 power (I am not used to this sites equation editor)

Thanks for your reply,
 
slider142 said:
Yes, your 1-parameter family of solutions contains every particular solution. In fact, every equation of this form, when solved by multiplying by the integrating factor of the exponential of the integral of p, will yield a 1-parameter family of solutions that contains every particular solution, as long as p and q are continuous on the interval of the solution.

Thank you for the reply.
 
Mark44 said:
The above isn't right. The left side is [y * (x + a)-2]', or in another form d/dx[y * (x + a)-2].
MACH2 said:
(x+a)^-2 = (x+a) elevated to the -2 power (I am not used to this sites equation editor)
It was the left side I was talking about, not the right side. Inside the brackets you should not have y'.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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