• • • ## ODE exponential solutions

\begin{aligned} \dot{x} = ax \qquad (1) \end{aligned}

\begin{aligned} \frac{dx}{dt} + p(t)x = g(t) \qquad (2) \end{aligned}

This is called the standard (canonical) form of the first order linear equation. Where both p(t) and g(t) are continuous functions (can be constants too).

Solution 1

Let's assume that the equation has an exponential solution (that is our guess):

\begin{aligned} x(t) = e^{ht} * x_0 \qquad (3) \end{aligned}

and plug it into the equation to find the value of h.

\begin{aligned} \dot{x} = ax \qquad \end{aligned}

Take derivative of the left side:

Since $e^{ht}x_0$ is never 0 we can divide both sides by it. Then:

\begin{aligned} h = a \qquad \end{aligned}

and assuming $x_0 \neq 0$.

The solution is then:

\begin{aligned} x(t) = e^{at}x_0 \qquad \end{aligned}

Solution 2 - variable separation

Assuming $x \neq 0$ divide both sides by $x$:

\begin{aligned}\frac{1}{x} \frac{dx}{dt} = a \qquad \end{aligned}

and multiply it by dt:

Now the variables are separated so we can integrate:

\begin{aligned} ln |x| = at + C \qquad \end{aligned}

Let both sides be exponents of the base e.

\begin{aligned} e^{ln |x|} = e^{at + C} \qquad \end{aligned}

Simplified:

\begin{aligned} |x| = e^{at + C} = e^C e^{at} = C_1 e^{at} \qquad \end{aligned}

Reference:

http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

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