I have the following differential equation: y+y=cos⁡(t)cos⁡(2t) Maybe something can be done to cos⁡(t)cos⁡(2t) to...



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I have the following differential equation:
Maybe something can be done to cos(t)cos(2t) to make it easier to solve. Any ideas?

Answer & Explanation



Beginner2021-11-24Added 19 answers

I don't see an elegant solution , but it can be done by brute force using the method of variation of parameters.
The homogeneous differential equation y(t)+y(t)=0 has the two fundamental solutions y1(t)=sin(t) and y2(t)=cos(t)
We must solve
for u1 and u2 and find U1 and U2 such that U1=u1 and U2=u2
A particular solution will then be given by
y0(t)=U1y1+U2y2 and y=y0+c1y1+c2y2
will be the general solution.
Since the matrix A(t)=(sintcostcostsint) is orthogonal, its inverse is its transpose, so
Note that v2(t)=sin(t)cos(t)cos(2t)=12sin(2t)cos(2t)=14sin(4t), so we can take
On other hand, we get

Charles Randolph

Charles Randolph

Beginner2021-11-25Added 16 answers

There's a general way of doing these things. You solve the homogeneous equation y+y=0, giving C1cos(t)sin(t)+C2sin(t) for constants C1 and C2, then add to it a single solution yp(t) to the inhomogeneous equation y+y=cos(t)cos(2t)=12cos(3t)+12cos(t). The resul will be the general solution to your differential equation
To find yp(t), you try yp(t)=a1cos(3t)+a2sin(3t)+b1tcos(t)+b2tsin(t). You plug it in and solve for a1,a2,b1, and b2. Normally you just try combinations of cos(3t),sin(3t),cos(t) and sin(t), but since the latter two solve the homogeneous equation you have to stick a t in front. I will trust Theo Buehler is right and that a1=116, b=14 and a2=b1=0. Thus your general solution will be
(The 516cos(t) term gets absorbed into the solution to the homogeneous equation).

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