1st order linear diff eq. problem

[Abraham]

Homework Statement

Find the general solution of

\frac{dy}{dt} = 2y +sin(2t)

Homework Equations

The general solution of a nonhomogeneous ode is the particular solution of the nonhomo plus the solution of the homogeneous ode.: y(t)= y_{p}(t)+y_{h}(t)

The Attempt at a Solution

\frac{dy}{dt} - 2y = sin(2t)


Initial guess, is that y_{p} = a*Sin(2t)+b*Cos(2t), where a and b are unknown constants later to be found

Then, the d.e. becomes:

\frac{d}{dt} ( a*Sin(2t)+b*Cos(2t) ) - 2( a*Sin(2t)+b*Cos(2t) ) = Sin2t

=2aCos(2t)-2bSin(2t) - 2aSin(2t)+2bCos(2t) = Sin2t

=(2a+2b)*Cos(2t) + (-2b-2a)*Sin(2t) = Sin2t

Using the fact that polynomial coefficients must be equal:

2a+2b=0
-2b-2a=1

And here is my issue. These simultaneous equations are false, and thus I cannot possibly solve the d.e.

Can someone tell me where I messed up? Thanks

PS. I know this can also be found using integrating factors, but i need help on the section with "guessing" the solution

 

[vela]

You got a sign wrong when you went from this line

\frac{d}{dt} (a \sin 2t + b \cos 2t ) - 2(a \sin 2t + b \cos 2t ) = \sin 2t

to your next one. You should have gotten

2a\cos 2t - 2b \sin 2t - 2a \sin 2t - 2b \cos 2t = \sin 2t

 

[Abraham]

Hey thanks. You homework helpers save my life.