Solving an ODE: Seeking c so y'(0)=0

[mariask]

so I am trying to solve this equation y'+5.6y=9.5cos(2x)+2.4sin(2x) . I want the c in order to y'(0)=0. I am really lost

 

[topsquark]

Do you know how to do the method of undetermined coefficients? Please let us know what you've done.

-Dan

 

[HOI]

so I am trying to solve this equation y'+5.6y=9.5cos(2x)+2.4sin(2x) . I want the c in order to y'(0)=0. I am really lost

There is NO "c" in what you wrote! Do you mean the constant in the solution? It isn't necessarily called "c"!

The general solution to the associated homogenous equation, y'+5.6y= 0, is \(\displaystyle y= Ce^{-5.6x}\). To find a solution to the entire equation, let \(\displaystyle y= Acos(2x)+ Bsin(2x)\). Then \(\displaystyle y'= -2Asin(2x)+ 2Bcos(2X)\) and the \(\displaystyle y'+ 5.6y= -2Asin(2x)+ 2Bcos(2x)+ 5.6Acos(2x)+5.6Bsin(2x)= (-2A+ 5.6B)sin(2x)+ (2B+ 5.6A)cos(2x)= 2.4sin(2x)+ 9.5cos(2x)\).

Since this to be true for all x, we must hav -2A+ 5.6B= 2.4 and 2B+ 5.6A= 9.5.

Solve those two equations for A and B. Then the solution to the entire equation is \(\displaystyle y= Ce^{-5.6x}+ Acos(2x)+ Bsin(2x)\) for those A and B. Then \(\displaystyle y'= -0.56Ce^{-5.6x}- 2Asin(2x)+ 2Bcos(2x)\) so that \(\displaystyle y'(0)= -0.56C+ 2B= 0\). \(\displaystyle C= (2/0.56)B\) where B was found before.