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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

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  • $\begingroup$ $a\ne0$ is not possible. $\endgroup$ – Yves Daoust Jun 7 at 16:15
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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 $1,4,5$.

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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.

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

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