I have a true-or-false question, and I need to find if the general solution to y''+y=0 is y(x)=c1(cosx)+5c2(cosx)
Apparently the answer is supposed to be false, but I don't understand why and have attempted to solve the problem multiple times.
Captain Matticus, LandPiratesInc
Favorite Answer
y = c[1] * cos(x) + 5c[2] * cos(x)
y = (c[1] + 5c[2]) * cos(x)
y' = -(c[1] + 5c[2]) * sin(x)
y'' = -(c[1] + 5c[2]) * cos(x)
If we say that c[1] + 5 * c[2] = c, then we have:
y = c * cos(x)
y'' = -c * cos(x)
y'' + y = -c * cos(x) + c * cos(x) = 0
So yes, it is a solution, but that doesn't mean that it's the general solution. That's the difference.
y'' + y = 0
The general solution is y = c[1] * sin(x) + c[2] * cos(x). Hopefully, you can see that the specific solution you were given has c[1] equal to 0 and c[2] equal to c + 5C
Anonymous
y(x) = sinx + cosx is a solution because
y'(x) = cosx - sinx, y''(x) = -sinx - cosx,
giving y'' + y = -sinx - cosx + sinx + cosx = 0.
But y(x) = sinx + cosx in not included in
c1(cosx) + 5c2(cosx) for any choice of c1, c2.
So answer is false.
General solution is y(x) = c1cosx + c2sinx.