Decide whether the method of undetermined coefficients together with superposition can be applied to find a particular solution of the given equation. Do not solve the equation.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{cosht}\mathbf{+}{\mathbf{sin}}^3\mathbf{t}$

Given equation,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{cosht}\mathbf{+}{\mathbf{sin}}^3\mathbf{t}$

Here, the given differential equation is non-homogeneous.

According to the method of undetermined coefficients,

The method of undetermined coefficients applies only to non-homogeneities that are polynomials, exponentials, sines or cosines, or products of these functions.

One gets, that the left-hand side consists of the differential equation with constants coefficients. So, there is no problem.

And the right-hand side mathematical expression is a linear combination of exponential and trigonometric functions.

Let${\mathrm{y}}_1$ be a solution of the differential equation,$\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}={\mathrm{f}}_1\left(\mathrm{t}\right)$ and${\mathrm{y}}_2$ be a solution of the differential equation, $\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}={\mathrm{f}}_2\left(\mathrm{t}\right)$.

Then for any constants${\mathrm{k}}_1$ and ${{\mathrm{k}}_2}_{}$, the function k,${\mathrm{k}}_1{\mathrm{y}}_1+{\mathrm{k}}_2{\mathrm{y}}_2$ is a solution to the differential equation,

$\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}={\mathrm{k}}_1{\mathrm{f}}_1\left(\mathrm{t}\right)+{\mathrm{k}}_2{\mathrm{f}}_2\left(\mathrm{t}\right)\dots \left(1\right)$

Given equation,

$\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+3\mathrm{y}=\mathrm{cosht}+{\sin }^3\mathrm{t}\dots \left(2\right)$

One knows that,

role="math" localid="1654939135196" $\begin{array}{c}\mathrm{cosht}=\frac{{\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}}{2}\\ {\sin }^3\mathrm{t}=\frac{3}{4}\mathrm{sint}-\frac{1}{4}\sin 3\mathrm{t}\end{array}$

Substitute the above formula in the equation (2),

role="math" localid="1654939198146" $\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+3\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}}{2}+\frac{3}{4}\mathrm{sint}-\frac{1}{4}\sin 3\mathrm{t}\dots \left(3\right)$

Compare equations (1) and (3),

role="math" localid="1654939257261" $\begin{array}{c}{\mathrm{y}}_1=\frac{{\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}}{2}\\ {\mathrm{y}}_2=\frac{3}{4}\mathrm{sint}-\frac{1}{4}\sin 3\mathrm{t}\end{array}$

So, the method of undetermined coefficients together with superposition can be applied.