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.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

So, the solution of the given equation is $\mathrm{y}={\mathrm{ce}}^{-2\mathrm{x}};\text{for}-\infty <\mathrm{x}<\infty \text{.}$

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 differential equation

$\frac{dy}{dx}+2y=0......\text{(1)}$

The standard form of the linear differential equation

$\frac{dy}{dx}+P\left(x\right)y=f\left(x\right).$

Clearly, equation (1) is in the standard form.

Evaluation

Compare the given differential equation with standard form, and identity p(x) =0 and f(x) = 0

The functions P=2 and f=0 are continuous on $(-\infty ,\infty )$

The integrating factor is

$$\begin{gathered} e^{\int P\left(x\right)dx}=e^{\int 2dx} \\ =e^{2x} \end{gathered}$$

Multiply the standard form, with the integrating factor

$$\begin{gathered} e^{2x}\frac{dy}{dx}+e^{2x}2y=0 \\ {e}^{2x}\frac{dy}{dx}+\frac{d}{dx}\left(e^{2x}\right)y=0 \\ \frac{d}{dx}\left(y{e}^{2x}\right)=0 \\ \int \frac{d}{dx}\left[e^{2x}y\right] \\ =cy \\ =c{e}^{-2x} \end{gathered}$$

Hence, the general solution of the given differential equation is$\mathrm{y}={\mathrm{ce}}^{-2\mathrm{x}};\text{for}-\infty <\mathrm{x}<\infty \text{.}$