Solving y'' + 5y' = 400sin(5t)+250cos(5t) with Intial Conditions

[mr_coffee]

This one got me alittle discombobulated. Initiallly I thought they wanted me to find a particular solution, which I did. But then later I saw it supplied to intial conditions. So I guessed y = Asin(5t)+Bcos(5t) then took derivatives, and applied the intial conditions but also was wrong. Here is the question:

Find the solution of
y'' + 5y' = 400sin(5 t) + 250cos(5 t)
with y(0) = 8 and y'(0) = 8 .
y =

Here is my work:
http://suprfile.com/src/1/1voo2y/lastscan.jpg

Thanks!

 

[TD]

Where is the solution of the homogenous part?

 

[HallsofIvy]

y= (8/5)sin(5t)+ 8 cos(5t) is NOT the general solution to the equation. You've left out the solution to the homogeneous part.

 

[assyrian_77]

mr_coffee, remember that it is $y=y_h+y_p$.

 

[mr_coffee]

O yeah i don't know what i was thinking, well I found the homogenous equation added it to the paricular, but still got it wrong:

For the homogenous i put:
y = A + Be^(-5t)
because the other r is 0, e^0 =1.

8 = y(0) = A+Be^(-5t);
8 = A+B;

8=y'(0) = -5Be^(-5t);
8 = -5B;
B = -8/5;

A = 8+8/5 = 48/5;

So for my answer i put:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/f7/03fa30705408dc77e802fec766ed1f1.png

Any ideas wehre i f'ed this one/

 

[assyrian_77]

The inital conditions apply to the total answer, i.e. write down $y=y_h+y_p$ and then apply the IC.

 

[mr_coffee]

I managed to not do it right again, i did what you said correctly i think...
here is my work:
http://suprfile.com/src/1/24jhk0/lastscan.jpg
Thanks!