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The actual task is to proof: $f''(x) + 25f(x) = 0$ then $f(x) = A\cos(5x) + B\sin(5x)$ for $A, B \in \mathbb{R}$

I tried to proof that for $g(x) = A\cos(5x)$ and $h(x) = B\sin(5x)$, but I can't get anywhere with that.

Regards,
Marco

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    What do you know about linear differential equations (or just with constant coefficients)? – 2017-01-06
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    Actually the solution in the title is not correct. It should be $A\cos(\sqrt{C}x)+B \sin(\sqrt{C}x)$. You also see this in your actual task since $\sqrt{C}=\sqrt{25}=5$. – 2017-01-06
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    @MarvinF you are right, I forgot to put that there. – 2017-01-06
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    @projectilemotion no, it is to proof that this is the only solution to the differential equation – 2017-01-06

5 Answers 5

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Consider the initial-value problem $$ f'' + Cf = 0,\qquad f(0) = A,\quad f'(0) = B\sqrt{C}. \tag{1} $$ Here's a sketch (parts of which you already know) of a proof that the unique solution of (1) is $$ f(x) = A\cos(\sqrt{C}x) + B\sin(\sqrt{C}x). $$ Throughout, "the ODE" refers to the differential equation $f'' + Cf = 0$ in (1).

  • If $f$ and $g$ are solutions of the ODE, so is every linear combination of $f$ and $g$.

  • The functions $c(x) = \cos(\sqrt{C}x)$ and $s(x) = \sin(\sqrt{C}x)$ are solutions of the ODE.

  • If $A = B = 0$, the unique soluton of (1) is the zero function.

    Proof: Suppose $g$ is a solution of (1) with $A = B = 0$. Since $0 = g'' + g$, multiplying by $2g'$ gives $$ 0 = 2g'\, g'' + 2g\, g' = \bigl[(g')^{2} + g^{2}\bigr]'. $$ By the identity theorem, the function in square brackets is constant, and by the initial conditions, this constant is $0$. In particular, $g = 0$ as a function.

  • The unique solution of (1) is $f(x) = A\cos(\sqrt{C}x) + B\sin(\sqrt{C}x)$.

    Proof: Let $f$ be an arbitrary solution of (1). The function $$ g(x) = f(x) - \bigl[A\cos(\sqrt{C}x) + B\sin(\sqrt{C}x)\bigr] $$ satisfies the ODE with initial conditions $g(0) = g'(0) = 0$.

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    Thank you! This is a good proof. – 2017-01-06
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    Do you know how I close this answer? – 2017-01-06
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    There should be a check box of some type next to each answer, maybe near the vote count...? – 2017-01-06
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    @AndrewD.Hwang I don't see how your proof of uniqueness works. Could you elaborate on it more? – 2017-01-06
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    @PiotrBenedysiuk: In the fourth step, $f$ is an arbitrary solution of (1); the difference $g$ between $f$ and the known solution solves the ODE with both initial conditions equal to $0$, so $g \equiv 0$. That is, $f$ itself must be the known solution. (There's a comparable proof of uniqueness for the first-order equation $f' = f$ and $f(0) = A$: Since $\exp$ solves the ODE, the function $g(x) = f(x)/e^{x}$ has vanishing derivative, and therefore is constant, i.e., $g(x) = g(0) = A$ for all $x$, or $f(x) = Ae^{x}$.) – 2017-01-06
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    I only now realize you specified the initial conditions. How would one go about proving all the solutions are part of the $span \{ \sin(\sqrt{C} x) , \cos(\sqrt{C} x) \}$ without stating the IC's? – 2017-01-06
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    @PiotrBenedysiuk: There are multiple approaches of greater or lesser generality. Offhand, for example, one could show that if $D$ represents the first derivative operator (acting on the space of $C^{2}$ functions) and if $k$ is complex, the operator $D + kI$ has one-dimensional kernel (as a complex vector space), so that $D^{2} + I = (D + iI)(D - iI)$ has $2$-dimensional kernel. Since your two functions are solutions and are not proportional over the complex numbers, they span the solution space. – 2017-01-06
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Hint: We know that by the chain rule of differentiation: $$\frac {d}{dx}(\cos ax) = a (-\sin ax) $$ and $$ \frac {d}{dx}(\sin ax)=a (\cos ax) $$ Can you take it from here?

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    I know that these are solutions to the differential equation. But I want to proof that these are the only solutions. – 2017-01-06
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$$f''(x)+Cf(x)=0$$

The procedure here is look for solutions of the form $ce^{\lambda x}$ i.e. we get by plugging into our equation and dividing by $ce^{\lambda x}$

$$\lambda^2+C=0$$

i.e. $\lambda=\pm i \sqrt{C}$. Now these two roots give you the solutions $f_1(x)=c_1 \exp(-i\sqrt{C}x)$ and $f_2(x)=c_2 \exp(i\sqrt{C}x)$. Now add these two solutions and use Euler's formula. You get

$$f(x)=(c_1+c_2) \cos(\sqrt{C}x)+i(-c_1+c_2) \sin(\sqrt{C}x)$$ And now just define $A=c_1+c_2$ and $B=i(c_2-c_1)$ to get the solution in your form.

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    But $B$ isn't $\in \mathbb{R}$, is it? – 2017-01-06
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First, note that both $g$ and $h$ are indeed solutions. (This is easy to see since second derivitive of sin or cos will give you itself back with a min sign.) Then, by linearity, sum of solutions is a solution as well. Thus $g+h$ is a solution.

Are you looking to prove that the span of $g,h$ are the only solutions?

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    I got that far. Yes that is what I am looking to proof – 2017-01-06
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    Are you working on some subspace of R or R itself? – 2017-01-06
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    on $\mathbb{R}$ itself. I thought maybe I can proof with $g(x) = f(x)\frac{1}{cos(x)}$ that $g'(x)$ is $0$. Which means $g(x)$ is constant. – 2017-01-06
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    You could show that $\sin(cx)$ and $\cos(cx)$ with $c \neq 5$ isn't an solution to the equation. Since cos and sines form a orthogonal basis the only solutions left are in span of $g$ and $h$ which is exactly of the form $f$ has. – 2017-01-06
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    Although, there must be a handy theorem here, but I'm not coming up with it. – 2017-01-06
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    but what about other functions than sin(cx) and cos(cx) – 2017-01-06
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One can use reduction of order. Let $f(x) = y(x)\cos{cx}$. Then $$ f''+ c^2 f = y''\cos{cx} + 2cy'\sin{cx}-c^2 y\cos{cx} + c^2 y\cos{cx} = y''\cos{cx} + 2cy'\sin{cx}. $$ This is a first-order differential equation for $y'$; we now look for the integrating factor, $ \exp{( \int 2c\tan{cx} \, dx )} = (\sec{cx})^2 $, and the equation becomes. $$ ( \cos^2{cx} \, y')' = 0 $$ The unique type of solution to $F'=0$ is a constant, so the unique way to integrate this equation is $$ y' = A\sec^2{cx}. $$ Again, this equation has a unique solution up to a constant, by direct integration: $$ y = A\tan{cx}+B, $$ and then $f(x) = A\sin{cx}+B\cos{cx}$. This works on any interval where $\cos{cx} \neq 0$, but we can refine it a bit to include the whole line by invoking continuity (or doing the same thing with $f(x)\sin{cx}$ and stitching together on intervals where they agree).