Question

Find the general solution of the given differential equation. Give the largest interval / over which the general solution is defined. Determine whether there are any transient terms in the general solution.

Short answer

So, the solution of the given equation is.$\mathrm{y}=\frac{1}{3}+{\mathrm{ce}}^{-4\mathrm{x}};$.

Step-by-step solution

Definition of transient term

A transient term means that you (or someone or something) will be moving on from where you are now.

Given data

Consider the following differential equation:

$3\frac{\mathrm{dy}}{\mathrm{dx}}+12\mathrm{y}=4$

Evaluation

Dividing by 3, we get the standard form

$\frac{\mathrm{dy}}{\mathrm{dx}}+4\mathrm{y}=\frac{4}{3}$

Multiplying the standard form by this factor, we get ${\mathrm{e}}^{4\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}+4{\mathrm{e}}^{4\mathrm{x}}\mathrm{y}=\frac{4}{3}{\mathrm{e}}^{4\mathrm{x}}$, which is same as $\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{e}}^{4\mathrm{x}}\mathrm{y}\right]=\frac{4}{3}{\mathrm{e}}^{4\mathrm{x}}$

On integration, we obtain

$$\begin{gathered} e^{4x}y=\frac{1}{3}{e}^{4x}+c \\ y=\frac{1}{3}+c{e}^{-4x};orfor-\infty <x<\infty \end{gathered}$$

${\mathrm{ce}}^{-4\mathrm{x}}$is a transient term, since, $e^{-4x}\to 0 asx\to \infty$.