- #1

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I've been banging my head with this one for a few days now. I've done (a) and (b) but I'm yet to do (c) if someone could complete this for me or help me along the way that would be greatly appreciated.

Thanks heaps!

- Thread starter Clef
- Start date

- #1

- 25

- 0

I've been banging my head with this one for a few days now. I've done (a) and (b) but I'm yet to do (c) if someone could complete this for me or help me along the way that would be greatly appreciated.

Thanks heaps!

- #2

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Suggestion: Solve the homogemous, find a particular solution, then i = i_t + i_p where i_t = Solution to teh homogenous, and i_p = particular solution.

Alternatively, if you knew Laplace Transform, you could try that as well.

- #3

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I've found my homogenous solution to be

(e^-3t)(Acos4t + Bsin4t)

I'm now attempting the particular solution by changing the original equation into the complex form:

z'' + 6z' +25z = -292e^(i4t)

And solving that. To be honest I am completely stuck where to go from here and my attempts have been going in a very roundabout way.

:(

- #4

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http://en.wikipedia.org/wiki/Method_of_variation_of_parameters

basically if {u1,u2} are the basis of the homogenous solution space, you assume the private solution

of the NH equation is:

A(t)*u1(t)+B(t)*u2(t)

playing around with the ODE you get the linear system

u1*(dA/dt)+u2*(dB/dt)=0

(du1/dt)*(dA/dt)+(du2/dt)*(dB/dt)=q(t)

where q(t) is the NH part. you get A' and B' and then by integration you get A(t),B(t).

Notice that A*u1+B*u2 will be the general solution for the ODE because A,B will contain

the free parameters (due to integration).

- #5

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I'm really sorry. But I do not completely understand this method. Would it be the same as substituting

http://en.wikipedia.org/wiki/Method_of_variation_of_parameters

basically if {u1,u2} are the basis of the homogenous solution space, you assume the private solution

of the NH equation is:

A(t)*u1(t)+B(t)*u2(t)

playing around with the ODE you get the linear system

u1*(dA/dt)+u2*(dB/dt)=0

(du1/dt)*(dA/dt)+(du2/dt)*(dB/dt)=q(t)

where q(t) is the NH part. you get A' and B' and then by integration you get A(t),B(t).

Notice that A*u1+B*u2 will be the general solution for the ODE because A,B will contain

the free parameters (due to integration).

y= tAcos4t+tbsin4t into the equation and then equating to get the PS?

- #6

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i= Acos4t + Bcos4t

for the solution for the particular solution of the whole DE?

Because I read somewhere that for some reason you have to put

i= t(Acos4t + Bsin4t)

but this is very very confusing :'(

- #7

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Variation of Parameters method is pretty simple once you have the basis of the solution-space.

The technique is very simple (without going into developing the method):

if you have a basis [tex]u_{j}[/tex] (in this case you found 2 functions)

then you get n linear equations of n variable [tex]k^{'}_{j}[/tex].

each equation is of the form:

[tex]\sum^{n}_{j=1}(\frac{d^{i}u_{j}}{dt^{i}}*k^{'}_{j})=0, \forall i=0,...,n-2[/tex]

and for i=n-1 you get the equation:

[tex]\sum^{n}_{j=1}(\frac{d^{n-1}u_{j}}{dt^{n-1}}*k^{'}_{j})=g(t)[/tex]

these are algebrical euqations. after you solve them you must integrate and then you

get n functions:

[tex]k_{j}(t)=\int(k^{'}_{j})dt+c_{j}[/tex]

and you put them back in the "linear"-combination

[tex]u_{general}(t)=\sum^{n}_{j=1}(k_j(t)*u_{j}(t))[/tex]

- #8

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okay. I think i've sort of got a handle on this way.

Would it be

=A'cos4t +B'sin4t = 0

-A'4sin4t + B'cos4t = -292sin4t

Then by equating coefficients and integrating

=A' = 73

therefore A = -73t

B' = 0

therefore B= 0

therefore the PS is -73tsin4t????

Would it be

=A'cos4t +B'sin4t = 0

-A'4sin4t + B'cos4t = -292sin4t

Then by equating coefficients and integrating

=A' = 73

therefore A = -73t

B' = 0

therefore B= 0

therefore the PS is -73tsin4t????

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