Question

Find the general solution of the following differential equations (complementary function particular solution). Find the solution by inspection or by (6.18), (6.23), or (6.24). Also find a computer solution and reconcile differences if necessary, noticing especially whether the solution is in simplest form [see (6.26) and the discussion after (6.15)].

$\mathbf{(}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{2}{\mathit{e}}^{\mathbf{x}}$

Short answer

The general solution of the differential equation is.$y=y_c+y_p={\alpha }_1{e}^{-ix}+{\alpha }_2{e}^{ix}+e^x$

Step-by-step solution

Given information 

A differential equation is given as.$(D^2-2D+1)y=0$

Auxiliary equation 

-Auxiliary equation:

Auxiliary equation is an algebraic equation of degreenupon which depends the solution of a given nth-order differential equation or difference equation.

-General form of the auxiliary equation $\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathbf{=}\mathit{k}{\mathit{e}}^{\mathbf{c}\mathbf{x}}$

Roots of the auxiliary equation

By guessing the particular- solution $y_p$for this differential equation, which is $y_p=e^x$and it is easy to prove that this is really a solution for this differential equation. Now to find the particular- solution for this differential equation by solving it have zero right-hand side, that is

$\begin{array}{c}(D^2+1)y=0\\ (D+i)(D-i)y=0\end{array}$

Notice that, the roots of the auxiliary equation are complex, and not equal, so the solution would be in the form of eq. (5.l6). Now, any solution for the equation $(D+i)y=0$is also a solution to our differential equation (see the textbook page 409 for more details).

$\begin{array}{c}{y}^{\text{'}}+iy=0\\ \frac{dy}{dx}=-iy\\ \int \frac{dy}{y}=-i\int dx\\ y={\alpha }_1{e}^{-ix}\end{array}$

General solution of differential equation 

By doing the same thing for the other simple equation, $(D-i)y=0$and get

$y={\alpha }_2{e}^{ix}$

So, the complementary solution is

$y_c={\alpha }_1{e}^{-ix}+{\alpha }_2{e}^{ix}$

And therefore, the general solution of the differential equation$(D^2+1)y=0$ is

$y=y_c+y_p={\alpha }_1{e}^{-ix}+{\alpha }_2{e}^{ix}+e^x$

By computer software write Simplify $\left[D\mathrm{Sol}ve\left[y,,\left[\text{x}\right]+\text{y}\left[\text{x}\right]==2{\text{E}}^{\wedge }\text{x},\text{y}\left[\text{x}\right],\text{x}\right]\right]$will get

$y=e^x+c_1\cos x+c_2\sin x$

But rewrite this result (i.e., the complementary solution) if write the exponential in terms of sine and cosine, and end up with the same result.