The purpose of this problem is to guide you through the process of finding the general solution of the linear non-homogeneous equation

$\displaystyle {y}{''}-{3}{y}'+{2}{y}={5}{e}^{{{3}{x}}}$

given that $\displaystyle {y}_{{1}}={e}^{{x}}$ is a solution of the associated homogeneous equation

$\displaystyle {y}{''}-{3}{y}'+{2}{y}={0}$.

First, use the reduction-of-order substitution $\displaystyle {y}={u}{y}_{{1}}$ to find a solution to the homogeneous equation

$\displaystyle {y}{''}-{3}{y}'+{2}{y}={0}$

Write your solution without coefficients or constants of integration.

$\displaystyle {y}_{{2}}=$ Preview Question 6 Part 1 of 3  

Now use the same substitution on the non-homogeneous equation

$\displaystyle {y}{''}-{3}{y}'+{2}{y}={5}{e}^{{{3}{x}}}$.

The result should be a first-order linear equation

$\displaystyle {w}'+{P}{\left({x}\right)}{w}={f{{\left({x}\right)}}}$

Use an integrating factor to solve this equation, and ultimately produce a particular solution to the non-homogeneous equation.

$\displaystyle {y}_{{p}}=$ Preview Question 6 Part 2 of 3  

Finally, write the general solution to the non-homogeneous equation in the form $\displaystyle {y}={c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}+{y}_{{p}}$. Use $\displaystyle {c}_{{1}}$ and $\displaystyle {c}_{{2}}$ as constants.

$\displaystyle {y}=$ Preview Question 6 Part 3 of 3