The discussion revolves around solving the differential equation y' - 2y/(x + a) = -1, where a is a constant. Participants are exploring the use of an integrating factor to find a solution.
There is an ongoing exploration of the solution's validity, with some participants affirming the existence of a 1-parameter family of solutions. However, others raise concerns about the differentiation steps and the formulation of the integrating factor.
Participants are discussing the implications of continuity for the functions involved in the differential equation and the assumptions related to the integrating factor used.
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
y = C(x+a)^2 + (x+a) this seems to be a solution, C = arbitrary constant
Is this the only solution??
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: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
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].
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.
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].
It was the left side I was talking about, not the right side. Inside the brackets you should not have y'.MACH2 said:(x+a)^-2 = (x+a) elevated to the -2 power (I am not used to this sites equation editor)