Personally, I would consider 6 easier than 5!
Problem 6 is a "linear equation with constant coefficients" and there is a standard method for solving it. The "associated homogeneous equation" is
dx2d2y+3dxdy+2y=0. It's "characteristic equation" is
r2+3r+2=(r+1)(r+2)=0 so its "characteristic roots" are -1 and -2. That tells us that the general solution to the associated homogeneous equation is
Ce−x+C2e−2x.
Now, to find the general solution to the entire equation, we just need to add any function that satisfies the entire equation. And recalling that the derivatives of sine and cosine are always again sine and cosine, I try a solution of the form
y=Asin(2x)+Bcos(2x) where I want to determine numbers A and B.
With
y=Asin(2x)+Bcos(2x),
dxdy=2Acos(2x)−2Bsin(2x) and
dx2d2y=−4Asin(2x)−4Bcos(2x).
So the equation becomes
dx2d2y+3dxdy+2y=−4Asin(2x)−4Bcos(2x)+6Acos(2x)−6Bsin(2x)+2Asin(2x)+2Bcos(2x)=(−4A−6B+2A)sin(2x)+(−4B+6A+2B)cos(2x)=(−2A−6B)sin(2x)+(−2B+6A)cos(2x)=10cos(2x).
Since this is to be true for all x we must have -2A- 6B= 0 and -2B+ 6A= 10. Multiply the first equation by 2 to get -4A- 12B= 0. Multiply the second equation by 6 to get -12B+ 36A= 60. Subtract the first of those from the second to eliminate B and get 40A= 60 so A= 6/4= 3/2. The -4A- 12B= -6- 12B= 0. -12B= 6 so B= -6/12= -1/2.
The general solution to the entire equation is
y(x)=C1e−x+C2e−2x+23sin(2x)−21cos(2x). The last thing to do, of course, is to determine
C1 and
C2 so that y(0)= 1 and y'(0)= 0.
y(0)=C1+C2+21=1.
y′(x)=−C1e−x−2C2e−x+3cos(2x)+sin(2x) so
y′(0)=−C1−2C2+3=0.