# Additional solutions to a given solution to a differential equation

It's my first week studying differential equations, and I would love some feedback regarding the following question:

A second order homogenous linear differential equation with constant coefficients has solution $$cos x$$. Which amongst the following functions have to also be solutions to our function?

$$f_1(x) = \frac{cos x}{3}$$

$$f_2(x) = 1$$

$$f_3(x) = cos x · sin x$$

$$f_4(x) = 3 sin x - 2 cos x$$

$$f_5(x) = cos (x + 2)$$

$$f_6(x) = 0.5x^2 + sin x$$

I have written down: 1, 4 and 5, as they are all versions of the same basic homogenous solution $$y = e^{ax}(c_1cosbx+c_2sinbx)$$. I am just worried I might be missing out on something, as often happens with these kinds of questions.

Many thanks

• $a\ne0$ is not possible. – Yves Daoust Jun 7 at 16:15

As a solution is $$\cos x$$, the characteristic polynomial has the roots $$\pm i$$, and the ODE must be

$$y''+y=0.$$

The general solution is thus

$$C\cos x+S\sin x$$

compatible with $$1,4,5$$.

As you stated, the general solution to the constant coefficient second order homogeneous differential equation has the form: $$y = e^{ax}(c_1 \cos(bx) + c_2\sin(bx )$$
Therefore, $$f_1(x) = \frac{\cos(x)}{3}$$ fits into this form as well as $$f_4(x) = 3\sin(x) - 2\cos(x)$$ and $$f_5(x) = \cos(x+2) = \cos(2)\cos(x) - \sin(2)\sin(x)$$ However, $$f_2(x) = 1$$ also fit into the general form.. Suppose i have a differential equation: $$2y'' - 5y' = 0$$ Then as you see here, $$f(x) = 1$$ would satisfy this ODE since the derivative of a constant is 0.

• You didn't use the information given. – Yves Daoust Jun 7 at 16:18