Solutions to the differential equation $ax''+bx=c$.

Question

I have the following differential equation: $$a\cdot x''(t) + b\cdot x(t)=c$$ where $a$, $b$, and $c$ are positive real numbers.

The associate polynomial would be: $$ s^2+b/a=0$$ thus, $$s=\pm j \sqrt{\frac{b}{a}}= j \omega_{0} $$

It is well know that the solution of this differential equation would be in the form:

$$ x(t)=A_{0} \cdot \sin (\omega_{0} t) + A_{1}\cdot \cos (\omega_{0} t) + A_{2} $$

where $A_{0}$, $A_{1}$ and $A_{2}$ are real numbers. The second derivative is $$x''(t)=-\omega_{0}^2 \cdot [A_{0} \cdot \sin (\omega_{0} t) + A_{1}\cdot \cos (\omega_{0} t)]$$

However, if I replace this solution on the differential equation something interesting is happenning:

$$ a\cdot x''(t)+b \cdot x(t)= (b-a\cdot \omega_{0}^2 )\cdot [A_{0} \cdot \sin (\omega_{0} t) + A_{1}\cdot \cos (\omega_{0} t)] + b \cdot A_{2}=c$$

The term $b-a\cdot \omega_{0}^2$ is always zero. So the constants $A_{0}$ and $A_{1}$ are undefined for any initial condition. There are infinite solutions for this differential equation. What I'm doing or assuming wrong when I get this kind of statement?

Answer 1

What you wrote is correct as a way to find the value of $A_2$ from the equation (this is the non-homogeneous part of the equation, a constant).

The constants $A_0$ and $A_1$ remains arbitrary as they define the general solution of the homogeneous equation. (As the equation is of the second order, there is a double infinity of solutions.)

Now to exploit the initial condition, you write

$$ x_0=x(0)=A_{0} \cdot \sin (\omega_{0} 0) + A_{1}\cdot \cos (\omega_{0} 0) + A_{2} $$ and $$A_1=x_0-A_2.$$

$A_0$ requires an additional condition, such as the initial derivative.

Answer 2

Your equation becomes $A_2=\frac{c}{b}$, which means that the set of all solutions to your differential equation is the set

$$\{A_0\sin(\omega_0t) + A_1\cos(\omega_0t)| A_0, A_1\in\mathbb R\}$$

Indeed, for every pair of real numbers $A_0, A_1$, the function $A_0\sin(\omega_0t) + A_1\cos(\omega_0t)$ solves the differential equation. There are infinitely many solutions.

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To actually calculate a single solution, you also need the initial conditions, for example, $x(0)$ and $x'(0)$, before you can calculate it.