2nd order differential equation

[howsaboutno]

Hi i need some help with solving this equation2 d²y/dt² + dy/dt +10y = 3sin(9t) - 8e^-2t - 7

when y=0 dy/dx = 10 t=0

The bit i am not sure about is the -8e^-2t - 7 bit on the right side because i only know how to deal with 2nd order differential equations when they are of the form a d²y/dt² +b dy/dx +cy = r(t).

any help would be much appreciated.

 

[tiny-tim]

Hi howsaboutno!

2 d²y/dt² + dy/dt +10y = 3sin(9t) - 8e^-2t - 7

when y=0 dy/dx = 10 t=0
…
i only know how to deal with 2nd order differential equations when they are of the form a d²y/dt² +b dy/dx +cy = r(t).

(i assume you mean d²y/dt² +b dy/dt +cy = r(t))

Not following you … it is of that form, isn't it?

 

[csprof2000]

You need one particular solution for each additive term in the RHS. The homogeneous solution works the same. The general solution is the homogeneous solution plus one particular solution for each of these additive terms.

 

[howsaboutno]

i don't understand what you mean csprof. and tiny tim yes it is in that form but I am still stuck.

 

[tiny-tim]

… and tiny tim yes it is in that form but I am still stuck.

For 8e^-2t, the normal thing to try would be e^-2t times a polynomial in t.

 

[csprof2000]

This is not a HW forum, so I assume that you want to see this problem worked out. It hasn't been moved, so I'll just do it.

Start with the homogeneous solution.

2y'' + y' + 10y = 0

2s^2 + s + 10 = 0
s = [-1 +- sqrt(1 - 80)] / 4

Call these s1, s2. So we have
y_h = A*exp(s1*t) + B*exp(s2*t)

So far, so good... yes?

Next, we do each term in the RHS one at a time, looking for particular solutions. We'll use the method of UC. So...

for 3sin(9t) we guess y_p1 = C*sin(9*t) + D*cos(9*t)
Work out: y_p1' and y_p1'' and plug into the original ODE. Solve for C, D. Now you have one particular solution.

for -8*exp(-2*t) we guess y_p2 = E*exp(-2*t).
Work out: y_p2' and y_p2'' and plug into the original ODE. Solve for E. Now you have another particular solution.

for 7 we guess y_p3 = F.
Work out: y_p3'' and y_p3' and plug into the original ODE. Solve for F. Now you have the third and final particular solution.

The answer to the problem is:

y_general = y_homogenous + y_p1 + y_p2 + y_p3

 

[howsaboutno]

thanks csprof although i did know what you meant just after i posted my previous comment. was just the homogeneous phrase that got me a little confused.

 

[HallsofIvy]

The nice thing about linear equations (and why they are so much easier than non-linear equations) is that you can do "pieces" separately, then put them back together to get the whole solution. You can do this thinking of the right hand side as [math]3sin(9t) - 8e^{-2t} - 7[/math], using a yp of [math]Asin(9t)+ Bcos(9t)+ Ce^{-2t}+ D[/math] or looking at the three parts separately- 3sin(9t) with yp= Asin(9t)+ Bcos(t)- [math]-8e^{-2t}[/math] with [math]yp= Ae^{-2t}[/math- and -7 with yp= A - and then put them together. Either way you will get the same answer.