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.

$\mathbf{y}\text{'}\mathbf{=}\mathbf{2}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{5}$

Short answer

So, the solution of the given equation is $y=-\frac{1}{2}{x}^2-\frac{1}{2}x-\frac{11}{4}+c{e}^{2x}$.

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

The objective is to find the general solution of the differential equation, $y^{\text{'}}=2y+x^2+5$and find the interval, over which the general solution is defined, and determine whether there are any transient terms in the general solution.

Consider the differential equation,$y^{\text{'}}=2y+x^2+5$.

Which is in the standard form, $\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)$.

Hence, the integrating factor is,

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

Evaluation

Multiply the standard form (1) with integrating factor, then

$$\begin{gathered} e^{-2x}{y}^{\text{'}}-2e^{-2x}y=e^{-2x}{x}^2+5e^{-2x} \\ \frac{d}{dx}\left[e^{-2x}y\right]=e^{-2x}{x}^2+5e^{-2x} \end{gathered}$$

Integrate on both sides, then

$$\begin{gathered} \int \frac{d}{dx}\left[e^{-2x}y\right]=\int \left(e^{-2x}{x}^2+5e^{-2x}\right)dx \\ {e}^{-2x}y=\int e^{-2x}{x}^{2}dx+5\int e^{-2x}dx \\ {e}^{-2x}y=\int e^{-2x}{x}^{2}dx-\frac{5}{2}{e}^{-2x}+C\dots \dots \left(2\right) \end{gathered}$$

Integration by Parts

Since, by product rule in integrals,$\int udv=uv-\int vdu$.

Consider, .

Let,$u=x^{2}anddv=e^{-2x}$, then

$\int e^{-2x}{x}^{2}dx=\left(x^2\right)\left(-\frac{1}{2}{e}^{-2x}\right)-\int \left(-\frac{1}{2}{e}^{-2x}\right)\left(2x\right)dx$

Finding transient terms

Substitute the value of $\int e^{-2x}{x}^{2}dx$in (2), then

$$\begin{gathered} e^{-2x}y=-\frac{1}{2}{x}^2{e}^{-2x}-\frac{x}{2}{e}^{-2x}-\frac{1}{4}{e}^{-2x}-\frac{5}{2}{e}^{-2x}+C \\ =-\frac{x^2}{2}{e}^{-2x}-\frac{x}{2}{e}^{-2x}-\frac{11}{4}{e}^{-2x}+C \end{gathered}$$

multiply e^2xwith on both sides, then

$y=-\frac{1}{2}{x}^2-\frac{1}{2}x-\frac{11}{4}+c{e}^{2x}$

Therefore, the general solution of the differential equation $y^{\text{'}}=2y+x^2+5$is

$y=-\frac{1}{2}{x}^2-\frac{1}{2}x-\frac{11}{4}+c{e}^{2x} for-\infty <x<\infty$

The general solution is defined over the interval $I:-\infty <x<\infty$.

Since, all terms in the general solution are not becomes zero as tends to infinity.

Therefore, there are no transient terms in the general solution.