# Prove that if $f''(x)+25f(x)=0$ then $f(x)=Acos(5x)+Bsin(5x)$ for some constants $A,B$

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is twice differentiable.

Prove that if $f''(x)+25f(x)=0$ then $f(x)=Acos(5x)+Bsin(5x)$ for some constants $A,B$

Consider $g(x):= f(x)-Acos(5x)-Bsin(5x)$ where $A$ and $B$ are chosen so that $g(0)=g'(0)=0$ [this portion is given, so can someone explain why it is important?]

For reference, we note that $g(x):= f(x)-Acos(5x)-Bsin(5x)$ $g'(x):= f'(x)+5Asin(5x)-5Bcos(5x)$ $g''(x):= f''(x)+25Acos(5x)+25Bsin(5x)$

Consider the derivative of $\frac{25}{2}g(x)^2+\frac{1}{2}g'(x)^2$

$25g(x)g'(x)+g''(x)g'(x)$ if and only if $g'(x)[25g(x)+g''(x)]$

Following substitution, we conclude that

$g'(x)[25g(x)+g''(x)]=g'(x)(0)=0$

Therefore, by the Constancy Theorem, $\frac{25}{2}g(x)^2+\frac{1}{2}g'(x)^2=C_0$, where $C_0$ is a constant

$25g(x)^2+g'(x)^2=C$, where $C$ is a constant

Thus,

$25(f(x)-Acos(5x)-Bsin(5x))^2 + (f'(x)+5Asin(5x)-5Bcos(5x))^2=C$

Where do I go from here?

• Note that $cos(5x)$ and $sin(5x)$ are solutions of your second order equation. and any solution will be linear combination of these solutions. – Arpit Kansal Apr 22 '15 at 18:27
• @ArpitKansal: That's what has to be proved, I think. – TonyK Apr 22 '15 at 18:35

By steps: $g(0)=g'(0)=0$ is important because you have a homogenous differential equation, therefore zero initial data implies that the solution is zero everywhere.

After obtaining that $25(f(x)-A\cos(5x)-B\sin(5x))^2 + (f'(x)+5A\sin(5x)-5B\cos(5x))^2=C$, use initial data $g(0)=g'(0)=0$.

We chose $A$ and $B$ to get $f(x)-A\cos(5x)-B\sin(5x)=0$ and $f'(x)+5A\sin(5x)-5B\cos(5x)=0$ at $x=0$. Therefore we can find $C$ by putting $x=0$: $$25(f(x)-A\cos(5x)-B\sin(5x))^2 + (f'(x)+5A\sin(5x)-5B\cos(5x))^2=C = 25(f(0)-A\cos(0)-B\sin(0))^2 + (f'(0)+5A\sin(0)-5B\cos(0))^2=0.$$

Hence $$\forall x\quad 25(f(x)-A\cos(5x)-B\sin(5x))^2 + (f'(x)+5A\sin(5x)-5B\cos(5x))^2 =0.$$ You have the sum of squares equal to zero, therefore each term is zero itself. We can conclude that $\forall x\, (f(x)-A\cos(5x)-B\sin(5x))^2=0$ and, therefore, $$\forall x\quad f(x)=A\cos(5x)+B\sin(5x).$$

We conclude that all solutions of the initial differential equation has the above form.

Since this is a differential equation with a right side equal to zero, we say that it is homogeneous. Thus, we solve the auxiliary equation associated with the equation:
f′′(x) + 25 f(x) = 0 ------> r^2 + 25 = 0

This yields:
r = +/- 5i

Solution will be of the form:
f = Ae^(5i) + be^(-5i)

Which using Euler's method yields:
f = ACos(5t) + BSin(5t).

Differentiating f twice and plugging into the original diff. eq. should yield zero.

Is this what you wanted?

I don't remenber the general formula, you can find in some book of ODE. You have this y= f(x)=> y'' + 25y=0. We make the substitution $y=\exp{rx}$. Then $r^2e^{rx}+25e^{}rx=0$ and $e^{rx}(r^2+25)=0$. We find the roots of polinomial $r^2+25$ there are $r=5i$ and $r=-5i$ for the properties of the solutions of the ode we can say $y = (e^{5xi}+e^-{-5xi})/2=cos(5x)$ is a solution or $y = (e^{5xi}+e^-{-5xi})/2=i sin(5x)$

In your expression involving $\frac {25}2g^2+\frac 12 g'^2=C$ you can set $x=0$ to determine the constant (this is where the bit you have been given is important). Then you have the sum of two squares equal to [?].

Note: you get the first part - the values of $A$ and $B$ from the values of $f(0)$ and $f'(0)$ if these are given, because $f(0)=g(0)+A$ so you want $A=f(0)$ and $f'(0)=g'(0)+5B$ so that $B=\frac 15 f'(0)$. Hence you can always find $A$ and $B$ which work.

If the initial conditions are given by the values of $f(x)$ at two distinct values of $x$ you get $A$ and $B$ by solving a couple of simultaneous equations.

Once you have filled in the details, this gives a proof that all the solutions of the original equation are of the requisite form, and justifies using that form to find solutions without replicating the proof every time.

Is it easier just to substitute? $$f(x)=Acos(5x)+Bsin(5x)$$ then $$\dot f(x)=-5Asin(5x)+5Bcos(5x)$$ $$\ddot f(x)=-25Acos(5x)-25Bsin(5x)$$ finally: $$-25Acos(5x)-25Bsin(5x)+25( Acos(5x)+Bsin(5x)) =0$$ ?

• I think the goal of this exercise is to show that there are no other solutions. – TZakrevskiy Apr 22 '15 at 18:28
• this only implies that f(x) is a solution. – Arpit Kansal Apr 22 '15 at 18:28