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# Give the correct answer and solve the given equation: displaystyle{y} text{ - 4y}+{3}{y}={x},{y}_{{1}}={e}^{x} # Give the correct answer and solve the given equation: displaystyle{y} text{ - 4y}+{3}{y}={x},{y}_{{1}}={e}^{x}

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Differential equations asked 2021-02-25
Give the correct answer and solve the given equation: $$\displaystyle{y}\ \text{ - 4y}+{3}{y}={x},{y}_{{1}}={e}^{x}$$

## Answers (1) 2021-02-26
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
$$\displaystyle{y}={C}_{{1}}{e}^{{{3}{x}}}+{C}_{{2}}{e}^{x}+\frac{1}{{3}}{x}+\frac{4}{{9}}$$

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