# differentialsWatch

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#1
A couple of issues withbdifferential equations

1.When the auxillary equation has repeated roots, you are left with one complimentary function so one constant of integration which doesnt follow linearly as it is 2nd order. So x (Ae^nx) is another standard one, which the book im using shows its true. But it doesnt explain why, can anyone here

2. For non homogeneous functions when you are finding the particular integral, you use a trial function that is of same form as RHS of differential. It does work obviously and I kinf of get why, but any explicit reasoning.

3.Also where the RHS is of the same form as the complimentary function the trial function is x times the trial function u would normally use for the RHS. I can see that using the trial function of same form of RHS insnt going to lead anywhere in thus situation. But why is it x timesd by it that works, why not something completely different or x^2 or something?
0
5 years ago
#2
(Original post by IceKidd)
A couple of issues withbdifferential equations

1.When the auxillary equation has repeated roots, you are left with one complimentary function so one constant of integration which doesnt follow linearly as it is 2nd order. So x (Ae^nx) is another standard one, which the book im using shows its true. But it doesnt explain why, can anyone here

2. For non homogeneous functions when you are finding the particular integral, you use a trial function that is of same form as RHS of differential. It does work obviously and I kinf of get why, but any explicit reasoning.

3.Also where the RHS is of the same form as the complimentary function the trial function is x times the trial function u would normally use for the RHS. I can see that using the trial function of same form of RHS insnt going to lead anywhere in thus situation. But why is it x timesd by it that works, why not something completely different or x^2 or something?
With 1. and 3., its about making the functions move complicated so they work. However we don't want to make them so overly complicated that they're a pain to solve. Multiplying by x is the simplest thing we can do to get them to work, which is essentially the methodology that goes through ones head when solving a differential equation: making something that works. You're right, if you multiplied your particular integral by x^2 it would probably work; its something to try out, however multiplying by x just makes it move simple.

with 2, you want something in the same form as the RHS that will satisfy the LHS. Lets say your RHS= sin(x) + 2cos(x), we wouldn't try using a polynomial, as that clearly wouldn't satisfy the RHS. This is the reason they are called 'trial' functions, as we are just trying different functions and seeing if they work. Again, with differential equations we want something that works.

If im completely wrong then im sorry, we're doing a whole module on differential equations in sixth form (MEI further maths) and this is the explanation that my teacher/book gave. You maybe doing much harder things at undergraduate that i haven't talked about. I hope i helped.
0
5 years ago
#3
(Original post by IceKidd)
A couple of issues withbdifferential equations

1.When the auxillary equation has repeated roots, you are left with one complimentary function so one constant of integration which doesnt follow linearly as it is 2nd order. So x (Ae^nx) is another standard one, which the book im using shows its true. But it doesnt explain why, can anyone here

2. For non homogeneous functions when you are finding the particular integral, you use a trial function that is of same form as RHS of differential. It does work obviously and I kinf of get why, but any explicit reasoning.

3.Also where the RHS is of the same form as the complimentary function the trial function is x times the trial function u would normally use for the RHS. I can see that using the trial function of same form of RHS insnt going to lead anywhere in thus situation. But why is it x timesd by it that works, why not something completely different or x^2 or something?
1.When you have the solution of the form then try solutions of the form . This should leave you with an equation in terms of , and in the example you stated this will be of the form

2. I'm not entirely sure about any reasoning behind this, I think you're just suppose to put something in which is of the same form with undetermined coefficients.

3. Do the same thing as what I mentioned for part 1, if you plug in a function multiplied by the trial function then you'll be able to see why you multiply by in this case.
0
#4
(Original post by ollz272)
With 1. and 3., its about making the functions move complicated so they work. However we don't want to make them so overly complicated that they're a pain to solve. Multiplying by x is the simplest thing we can do to get them to work, which is essentially the methodology that goes through ones head when solving a differential equation: making something that works. You're right, if you multiplied your particular integral by x^2 it would probably work; its something to try out, however multiplying by x just makes it move simple.

with 2, you want something in the same form as the RHS that will satisfy the LHS. Lets say your RHS= sin(x) + 2cos(x), we wouldn't try using a polynomial, as that clearly wouldn't satisfy the RHS. This is the reason they are called 'trial' functions, as we are just trying different functions and seeing if they work. Again, with differential equations we want something that works.

If im completely wrong then im sorry, we're doing a whole module on differential equations in sixth form (MEI further maths) and this is the explanation that my teacher/book gave. You maybe doing much harder things at undergraduate that i haven't talked about. I hope i helped.
(Original post by jameswhughes)
1.When you have the solution of the form then try solutions of the form . This should leave you with an equation in terms of , and in the example you stated this will be of the form

2. I'm not entirely sure about any reasoning behind this, I think you're just suppose to put something in which is of the same form with undetermined coefficients.

3. Do the same thing as what I mentioned for part 1, if you plug in a function multiplied by the trial function then you'll be able to see why you multiply by in this case.
Very helpful replies thank you both
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