# 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:
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The homogeneous equation is $y-4{y}^{\prime }+3y=0$
Its charecteristic equation is
${r}^{2}-4r+3=0⇒r=\frac{4±\sqrt{{\left(-4\right)}^{2}-4\cdot 1\cdot 3}}{2\cdot 1}=\frac{4±2}{2}=2±1$
Thus, the solutions of the characteristic equation are ${r}_{1}=3,{r}_{2}=1$, so the solution of (1) is
$y={C}_{1}{e}^{{r}_{1}x}+{C}_{2}{e}^{{r}_{2}x}={C}_{1}{e}^{3x}+{C}_{2}{e}^{x}$
where ${C}_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{C}_{2}$ are constants.
Now we need to find a particular solution, that is, we have to guess some function y such that $y-4{y}^{\prime }+3y=x\left(2\right)$
Since the RHS is a polynomial of degree 1, we will try with $y=ax+b.$
Then ${y}^{\prime }=a,y=0$
so (2) becomes
$-4a+3ax+3x=x⇒3ax+\left(3b-a\right)=1x$
From this we get
$3a=1$
$3b-4a=0$
Solving this system we get
$a=\frac{1}{3},3b=4a=\frac{4}{3}⇒b=\frac{4}{9}$
Therefore, the particular solution is
$y=ax+b=\frac{1}{3}x+\frac{4}{9}$
Finally, the solution of the original equation is given by the sum of the homogeneous and particular solution, so
$y={C}_{1}{e}^{3x}+{C}_{2}{e}^{x}+\frac{1}{3}x+\frac{4}{9}$