Find a particular solution to the differential equation. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{+}\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{2}}^x$

The given differential equation is:

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)=2^{\mathrm{x}}$

Simplify the above equation by using logarithms properties,

$\begin{array}{c}\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\ln \left(2^{\mathrm{x}}\right)}\\ \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}\left[\ln \left(2\right)\right]}\\ \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\left[\ln \left(2\right)\right]\mathrm{x}}\end{array}$

The auxiliary equation for the above equation:

${\mathrm{m}}^2+1=0$

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+1=0\\ {\mathrm{m}}^2=-1\\ \mathrm{m}=\sqrt{-1}\\ \mathrm{m}=\pm \mathrm{i}\end{array}$

The roots of the auxiliary equation are:

${\mathbf{m}}_1\mathbf{=}\mathbf{i}\&{\mathbf{m}}_2\mathbf{=}\mathbf{-}\mathbf{i}$

The complementary solution of the given equation is:

${\mathbf{y}}_c\mathbf{=}{\mathbf{c}}_1\mathbf{cos}\left(\mathbf{x}\right)\mathbf{+}{\mathbf{c}}_2\mathbf{sin}\left(\mathbf{x}\right)$

According to the method of undetermined coefficients, assume the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}={\mathrm{Ae}}^{\left[\ln \left(2\right)\right]\mathrm{x}}\dots \left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=\mathrm{Aln}\left(2\right){\mathrm{e}}^{\left[\ln \left(2\right)\right]\mathrm{x}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=\mathrm{A}{\left[\ln \left(2\right)\right]}^2{\mathrm{e}}^{\left[\ln \left(2\right)\right]\mathrm{x}}\end{array}$

From the equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+{\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{\left[\ln \left(2\right)\right]\mathrm{x}}\\ \mathrm{A}{\left[\ln \left(2\right)\right]}^2{\mathrm{e}}^{\left[\ln \left(2\right)\right]\mathrm{x}}+{\mathrm{Ae}}^{\left[\ln \left(2\right)\right]\mathrm{x}}={\mathrm{e}}^{\left[\ln \left(2\right)\right]\mathrm{x}}\\ {\mathrm{Ae}}^{\left[\ln \left(2\right)\right]\mathrm{x}}[{\left[\ln \left(2\right)\right]}^2+1]={\mathrm{e}}^{\left[\ln \left(2\right)\right]\mathrm{x}}\\ \mathrm{A}[{\left[\ln \left(2\right)\right]}^2+1]=1\\ \mathrm{A}=\frac{}{{\left[\ln \left(2\right)\right]}^2+1}\end{array}$

Substitute the value of A in the equation (2), we get:

${\mathrm{y}}_{\mathrm{p}}=2^{\mathrm{x}}{({\left[\ln \left(2\right)\right]}^2+1)}^{-1}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\left(\frac{1}{{\left[\ln \left(2\right)\right]}^2+1}\right){\mathrm{e}}^{\left[\ln \left(2\right)\right]\mathrm{x}}\\ {\mathrm{y}}_{\mathrm{p}}=\frac{2^{\mathrm{x}}}{{\left[\ln \left(2\right)\right]}^2+1}\\ {\mathrm{y}}_{\mathrm{p}}=2^{\mathrm{x}}{({\left[\ln \left(2\right)\right]}^2+1)}^{-1}\end{array}$