How Do You Determine When to Multiply by \(x\) or \(x^2\) in ODE Solutions?

  • Thread starter Thread starter brad sue
  • Start date Start date
  • Tags Tags
    General Ode
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
brad sue
Messages
270
Reaction score
0
Hi I have problem to answer this question. I think that it is a question of recognizing the right solution rather than computing.

Give the general solution of the differential equation:
y'' -6y'+9y=-5e(3x)+(1/x)e(3x)

1- y=C1*e(3x)+C2*x*e(3x)-(5/2)x2*e(3x)+x*e(3x)*ln(x)

2- y=C1*e(3x)+C2*x*e(3x)-(5/2)x*e(3x)-x*e(3x)*ln(x)

3-y=C1*e(3x)+C2*x*e(3x)-(5/2)*e(3x)+e(3x)*ln(x)

My problem is when to know if the particular solution is multiply by x or x^2.
when the particular solution is a solution of the homogenous equation , we multiply the particular solution by x. when do we multiply by x^2??
Can you give me a quick example please??
Thank you
B.
 
Physics news on Phys.org
you would multiply the paticular solution by x^2 when there is a repeated root of multiplicity of 3.
In other words if the solution had a triple root of 1.

a*e^x + b*x*e^x + c*x^2*e^x
 
brad sue said:
Hi I have problem to answer this question. I think that it is a question of recognizing the right solution rather than computing.

Give the general solution of the differential equation:
y'' -6y'+9y=-5e(3x)+(1/x)e(3x)

1- y=C1*e(3x)+C2*x*e(3x)-(5/2)x2*e(3x)+x*e(3x)*ln(x)

2- y=C1*e(3x)+C2*x*e(3x)-(5/2)x*e(3x)-x*e(3x)*ln(x)

3-y=C1*e(3x)+C2*x*e(3x)-(5/2)*e(3x)+e(3x)*ln(x)

My problem is when to know if the particular solution is multiply by x or x^2.
when the particular solution is a solution of the homogenous equation , we multiply the particular solution by x. when do we multiply by x^2??
Can you give me a quick example please??
Thank you
B.

It should be obvious that the correct answer can't be 2 or 3 because
-(5/2)*e(3x) could be absorbed into the C1*e(3x) term and (5/2)x*e(3x) could have been absorbed into C2*x*e(3x). Those don't give anything new.