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.

$\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}$

Short answer

So, the solution of the given equation is$y=\frac{1}{3}+c{e}^{-x^3};\text{for}-\infty <x<\infty \text{.}$

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 following differential equation:

$y\text{'}+3x^{2}y=x^2$

The standard form of the linear differential equation

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

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

Evaluation

Compare the given differential equation with standard form, and identity$P\left(x\right)=3x^2$ and$f\left(x\right)=x^2.$

The functions$P=3x^2$ and$f=x^2$are continuous on$(-\infty ,\infty )$.

The integrating factor is,

${\mathrm{e}}^{\int \mathrm{\mu }\left(\mathrm{x}\right)\mathrm{dx}}={\mathrm{e}}^{\int 3{\mathrm{x}}^2\mathrm{dx}}$

$=e^{3\left(\frac{x^3}{3}\right)}$

$=e^{x^3}$

Integrating the function

Multiply the standard form, with the integrating factor

$$\begin{gathered} e^{x^3}{y}^{\text{'}}+3x^2{e}^{x^3}y=x^2{e}^{x^3} \\ \frac{d}{dx}\left[e^{x^3}y\right]=x^2{e}^{x^3} \end{gathered}$$

$\text{Since}\frac{d}{dx}\left(x^3\right)=3x^2$

Integrate on both sides

$$\begin{gathered} e^{x^3}y=\int x^2{e}^{x^3}dx \\ =\frac{1}{3}\int 3x^2{e}^{x^3}dx \end{gathered}$$

Finding general equation

Substitute X³= t then $3{\mathrm{x}}^2\mathrm{dx}=\mathrm{dt}$

$e^{x^3}y=\frac{1}{3}\int e^{\text{'}}dt$

$=\frac{1}{3}{e}^t+c,c\text{isintegratingconstant}$

$=\frac{1}{3}{e}^{\left(x^3\right)}+c\text{Since}{x}^3=t$

$=\frac{e^{x^3}}{3}+c$

Therefore,$y\left(x\right)=\frac{1}{3}+c{e}^{-x^3}$

Hence the general solution of the given differential equation is $y=\frac{1}{3}+c{e}^{-x^3};\text{for}-\infty <x<\infty \text{.}$In general solution,$y_p=\frac{1}{3}and{y}_c=c{e}^{-x^3}$

For large values of x, the value of is negligible.

That is localid="1668415281027" $y_e\to 0asx\to \infty$,

Therefore, $y_e\left(x\right)=c{e}^{-x^3}$ is a transient term in the general solution.