2nd order nonhomogeneos differential equations with initial conditions

[pedro123]

Homework Statement

The problem states

d^2y/dt^2 +15y= cost4t + 2sin t

initial conditions y(0)=y'(0)=0

Homework Equations

The Attempt at a Solution

All I have is this r^2+15=0

making r(+-)=√15


and making yh= C1cos√15+C2√15


the next part includes solve for the nonhomogenous equation

cos4t + 2sint

but here is where I can't come with the particular solution can it be


yp= Acos2t +B sin2t

 

[Mark44]

Homework Statement

The problem states

d^2y/dt^2 +15y= cost4t + 2sin t

initial conditions y(0)=y'(0)=0

Homework Equations

The Attempt at a Solution

All I have is this r^2+15=0

making r(+-)=√15

No, r = ±√(-15) = ±i√(15).

But rather than working with e^it√15 and e^-it√15, it's much easier to work with cos(t√(15)) and sin(t√(15)).

and making yh= C1cos√15+C2√15

See above.

What would you have for y_h, considering the corrections I made?

the next part includes solve for the nonhomogenous equation

cos4t + 2sint

but here is where I can't come with the particular solution can it be


yp= Acos2t +B sin2t

No, these won't work. You'll need four expressions for your particular solution, not two.

 

[pedro123]

So yh= cos(t√(15)) + sin(t√(15))

and now I need to use the method of undetermined cooficients to come with a particular solution.

is Yp= Acos 2t +B sin 2t

Yp'=-2Asin 2t + 2B cos 2t
Yp''= -4Acos 2t - 4B sin 2t

right or did I make mistake and if it is right

it will make

d^2y/dt^2 +15Yp= -4Acos 2t - 4B sin 2t + 15 Acos 2t +15B sin 2t

d^2y/dt^2 +15Yp= 11cost + 11sint

is the particular solution right

 

[CAF123]

So yh= cos(t√(15)) + sin(t√(15))

That is still not the full general solution to the homogeneous equation.

and now I need to use the method of undetermined cooficients to come with a particular solution.

is Yp= Acos 2t +B sin 2t

Yp'=-2Asin 2t + 2B cos 2t
Yp''= -4Acos 2t - 4B sin 2t

right or did I make mistake and if it is right

it will make

d^2y/dt^2 +15Yp= -4Acos 2t - 4B sin 2t + 15 Acos 2t +15B sin 2t

Could you explain more what you are doing here? In particular why are substituting a form where the argument of sin is 2t?

 

[HallsofIvy]

Your non-homogeneous part has sin(t) and sin(4t). Those do NOT "combine" to give sin(2t).
Look for a solution of the form A sin(t)+ B cos(t)+ C sin(4t)+ D cos(4t).

 

[pedro123]

so you mean for the nonhomegenous part

yp=A sin(t)+ B cos(t)+ C sin(4t)+ D cos(4t).
yp'=A cos(t) - B sin(t) + 4C cos (4t) + - 4D sin (4t)
yp''= -A sin(t) - B cos(t) - 16C sin(4t) - 16D cos(4t)

 

[Mark44]

Yes. Now substitute y_p and its derivatives into your diff. equation to solve for A, B, C, and D.

For your general solution you will also need y_h, which wasn't right in post #3.

 

[pedro123]

so you mean Yp''-Yp'+15Yp=cost4t to solve

-A sin(t) - B cos(t) - 16C sin(4t) - 16D cos(4t) + 15A sin(t)+ 15B cos(t)+ 15C sin(4t)+ 15D cos(4t) = cos (4t ) + 2 sin(t)

-14 A sin(t) +14 B cos(t) -C sin(4t) -D cos(4t) = cos (4t) + 2 sin t

-C = 1 = -1

-D = 1 = -1

-14A = 1 = -1/14

14B = 1 = 1/14

are all these values for A B C and D right and my differential equation

 

[Mark44]

so you mean Yp''-Yp'+15Yp=cost4t to solve

-A sin(t) - B cos(t) - 16C sin(4t) - 16D cos(4t) + 15A sin(t)+ 15B cos(t)+ 15C sin(4t)+ 15D cos(4t) = cos (4t ) + 2 sin(t)

-14 A sin(t) +14 B cos(t) -C sin(4t) -D cos(4t) = cos (4t) + 2 sin t

-C = 1 = -1

-D = 1 = -1

Don't write things like the above. You're saying that -C is equal to 1 and that 1 is equal to -1, which isn't true. You should write something like this
-C = 1, so C = -1
or
-C = 1 $\Rightarrow$ C = -1

-14A = 1 = -1/14

14B = 1 = 1/14

are all these values for A B C and D right and my differential equation

I don't know - check them and see.

If y_p = (-1/14)sin(t) + (1/14)cos(t) - sin(4t) - cos(4t),
what is y_p'' + 15y_p? It better be equal to cos(4t) + 2sin(t).