Question

IGNORE (explaining something to a friend)

Answer 1

y''+64y=5t+32sin(8t)

Answer 2

these are the notes i have to go off of...

Answer 3

I understand how to get the Yc or the complimentary solution or whatever you call it that's easy

Answer 4

at least i hope Yc = C1Cos(8t)+C2Sin(8t)

Answer 5

Okay, so first things first. Solve the homogenous part.
\[\phi^2+64=0\]
From that you can deduce that:
\[\phi= \pm 8i\]
From this you know you have a homogenous solution of:
\[y_H(t)=c_1 \sin(8t)+ c_2 \cos(8t)\]
Now to solve your non-homogenous part. You see that the right side is something in the form:
\[At+C+Dsin(8t)+Ecos(8t)\]
Where, in this case, C and E would be 0.
Now, notice in your homogenous solution you have a "sin(8t)" this means you must multiply the part of your particular solution (in the form above^) by t. So it becomes:
\[At+C+Dtsin(8t)+Etcos(8t)\]
So you want your particular solution to be something of this form.
So set it to that:
\[y_P(t)=At+C+Dtsin(8t)+Etcos(8t)\]
Differentiate that twice and plug it into your diffeq.
Differentiating we find:
\[y'_P(t)=A+D(\sin(8t)+8tcos(8t))+E(\cos(8t)-8tsin(8t))\]
Differentiating again we find:
\[y''_P(t)=8Dcos(8t)+8D(\cos(8t)-8tsin(8t))-8Esin(8t)-8E(\sin(8t)+8tcos(8t))\]
Plugging this in the differential equation becomes:
\[8Dcos(8t)+8D(\cos(8t)-8tsin(8t))-8Esin(8t)-8E(\sin(8t)+8tcos(8t))\]
\[+64(AT+C+Dt \sin(8t)+Et \cos(8t)=5t+32\sin(8t)\]
Now you want to use your knowledge of linear algebra to find the constants. This must mean:
\[64A=5; 16D=0;-16E=32;-63D=0;C=0\]
So that makes your particular solution:
\[y_P(t)=\frac{5}{64}t-2t \cos(8t)\]
Adding this to the homogenous part, you get your solution:
\[y(t)=\frac{5}{64}t-2t \cos(8t)+c_1\cos(8t)+c_2\sin(8t)\]