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.

$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{t}\mathbf{+}{\mathbf{sin}}^2\mathbf{t}\mathbf{+}\mathbf{3}$

Given equation,

$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{t}\mathbf{+}{\mathbf{sin}}^2\mathbf{t}\mathbf{+}\mathbf{3}$

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, sine, cosine, 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 algebraic 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,

$2\mathrm{y}\text{'}\text{'}+3\mathrm{y}\text{'}-4\mathrm{y}=2\mathrm{t}+{\sin }^2\mathrm{t}+3\dots \left(2\right)$

One knows that,

role="math" localid="1654939894668" ${\sin }^2\mathrm{t}=\frac{1-\cos \left(2\mathrm{t}\right)}{2}$

Substitute the above formula in the equation (2),

role="math" localid="1654939959531" $\begin{array}{c}2\mathrm{y}\text{'}\text{'}+3\mathrm{y}\text{'}-4\mathrm{y}=2\mathrm{t}+\frac{1-\cos \left(2\mathrm{t}\right)}{2}+3\\ 2\mathrm{y}\text{'}\text{'}+3\mathrm{y}\text{'}-4\mathrm{y}=2\mathrm{t}+\frac{7}{2}-\frac{\cos \left(2\mathrm{t}\right)}{2}....\left(3\right)\end{array}$

Compare equations (1) and (3),

role="math" localid="1654940000526" $\begin{array}{c}{\mathrm{y}}_1=2\mathrm{t}+\frac{7}{2}\\ {\mathrm{y}}_2=\frac{\cos \left(2\mathrm{t}\right)}{2}\end{array}$

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