The homogeneous equation is \(\displaystyle{y}\text{}-{4}{y}'+{3}{y}={0}\)

Its charecteristic equation is

\(\displaystyle{r}^{2}-{4}{r}+{3}={0}\Rightarrow{r}=\frac{{{4}\pm\sqrt{{{\left(-{4}\right)}^{2}-{4}\cdot{1}\cdot{3}}}}}{{{2}\cdot{1}}}=\frac{{{4}\pm{2}}}{{2}}={2}\pm{1}\)

Thus, the solutions of the characteristic equation are \(\displaystyle{r}_{{1}}={3},{r}_{{2}}={1}\), so the solution of (1) is

\(\displaystyle{y}={C}_{{1}}{e}^{{{r}_{{1}}{x}}}+{C}_{{2}}{e}^{{{r}_{{2}}{x}}}={C}_{{1}}{e}^{{{3}{x}}}+{C}_{{2}}{e}^{x}\)

where \(\displaystyle{C}_{{1}}{\quad\text{and}\quad}{C}_{{2}}\) are constants.

Now we need to find a particular solution, that is, we have to guess some function y such that \displaystyle{y}\text{}-{4}{y}'+{3}{y}={x}{\left({2}\right)}

Since the RHS is a polynomial of degree 1, we will try with \(\displaystyle{y}={a}{x}+{b}.\)

Then \(\displaystyle{y}'={a},{y}\text{}={0}\)

so (2) becomes

\(\displaystyle-{4}{a}+{3}{a}{x}+{3}{x}={x}\Rightarrow{3}{a}{x}+{\left({3}{b}-{a}\right)}={1}{x}\)

From this we get

\(\displaystyle{3}{a}={1}\)

\(\displaystyle{3}{b}-{4}{a}={0}\)

Solving this system we get

\(\displaystyle{a}=\frac{1}{{3}},{3}{b}={4}{a}=\frac{4}{{3}}\Rightarrow{b}=\frac{4}{{9}}\)

Therefore, the particular solution is

\(\displaystyle{y}={a}{x}+{b}=\frac{1}{{3}}{x}+\frac{4}{{9}}\)

Finally, the solution of the original equation is given by the sum of the homogeneous and particular solution, so

\(\displaystyle{y}={C}_{{1}}{e}^{{{3}{x}}}+{C}_{{2}}{e}^{x}+\frac{1}{{3}}{x}+\frac{4}{{9}}\)

Its charecteristic equation is

\(\displaystyle{r}^{2}-{4}{r}+{3}={0}\Rightarrow{r}=\frac{{{4}\pm\sqrt{{{\left(-{4}\right)}^{2}-{4}\cdot{1}\cdot{3}}}}}{{{2}\cdot{1}}}=\frac{{{4}\pm{2}}}{{2}}={2}\pm{1}\)

Thus, the solutions of the characteristic equation are \(\displaystyle{r}_{{1}}={3},{r}_{{2}}={1}\), so the solution of (1) is

\(\displaystyle{y}={C}_{{1}}{e}^{{{r}_{{1}}{x}}}+{C}_{{2}}{e}^{{{r}_{{2}}{x}}}={C}_{{1}}{e}^{{{3}{x}}}+{C}_{{2}}{e}^{x}\)

where \(\displaystyle{C}_{{1}}{\quad\text{and}\quad}{C}_{{2}}\) are constants.

Now we need to find a particular solution, that is, we have to guess some function y such that \displaystyle{y}\text{}-{4}{y}'+{3}{y}={x}{\left({2}\right)}

Since the RHS is a polynomial of degree 1, we will try with \(\displaystyle{y}={a}{x}+{b}.\)

Then \(\displaystyle{y}'={a},{y}\text{}={0}\)

so (2) becomes

\(\displaystyle-{4}{a}+{3}{a}{x}+{3}{x}={x}\Rightarrow{3}{a}{x}+{\left({3}{b}-{a}\right)}={1}{x}\)

From this we get

\(\displaystyle{3}{a}={1}\)

\(\displaystyle{3}{b}-{4}{a}={0}\)

Solving this system we get

\(\displaystyle{a}=\frac{1}{{3}},{3}{b}={4}{a}=\frac{4}{{3}}\Rightarrow{b}=\frac{4}{{9}}\)

Therefore, the particular solution is

\(\displaystyle{y}={a}{x}+{b}=\frac{1}{{3}}{x}+\frac{4}{{9}}\)

Finally, the solution of the original equation is given by the sum of the homogeneous and particular solution, so

\(\displaystyle{y}={C}_{{1}}{e}^{{{3}{x}}}+{C}_{{2}}{e}^{x}+\frac{1}{{3}}{x}+\frac{4}{{9}}\)