Laplace transformations on a homogeneous ODE

Question

$$y^{\prime\prime} - 3y^{\prime} + 2y = 0$$ $y(0) = 14$, $y^{\prime}(0)=0$, and using the Laplace transformation I'm trying to solve this IVP

Answer 1

For $$y^{\prime\prime} - 3y^{\prime} + 2y = 0\tag{1}$$

Let $$Y=\mathscr{L}(y)=\int_{t=0}^{\infty}e^{-pt}f(t)\,\mathrm{d}t$$ where $f(t)=y$ and take the Laplace transform of both sides of $(1)$ term by term we get $$\color{red}{\mathscr{L}(y^{\prime\prime})=p^2Y-py(0)-{y^{\prime}(0)}}$$ and $$\color{red}{\mathscr{L}(3y^{\prime})=3pY-3y(0)}$$ and $$\mathscr{L}(2y)=2\cdot{\mathscr{L}(y)=2Y}$$ lastly $${\mathscr{L}(0)=0}$$

The equations marked $\color{red}{\mathrm{red}}$ can be obtained from the end of This Table of Laplace transforms

Now $(1)$ becomes (due to linearity of Laplace transforms) $$\mathscr{L}(y^{\prime\prime})-\mathscr{L}(3y^{\prime})+\mathscr{L}(2y)=p^2Y-py(0)-{y(0)}^{\prime}-3pY+3y(0)+2Y=0\tag{2}$$ Now substitute $y(0) = 14$ and $y^{\prime}(0)=0$ into $(2)$ to obtain $$p^2Y-14p-3pY-42+2Y=0$$ $$\implies Y(p^2-3p+2)=14(p+3)$$ $$\implies Y=\frac{14(p+3)}{(p-1)(p-2)}=\frac{14p}{(p-1)(p-2)}+\frac{42}{(p-1)(p-2)}$$

Now take the inverse Laplace of both sides $$\implies y= \mathscr{L}^{-1}(Y)=\mathscr{L}^{-1}\left(\frac{14p}{(p-1)(p-2)}\right)+\mathscr{L}^{-1}\left(\frac{42}{(p-1)(p-2)}\right)\tag{3}$$ by Linearity which is obeyed by Laplace and inverse Laplace transforms.

Now, in general obtained from the start of this Table of Laplace transformations: $$\mathscr{L}^{-1}\left(\frac{p}{(p+a)(p+b)}\right)=\frac{ae^{-at}-be^{-bt}}{a-b}$$ and $$\mathscr{L}^{-1}\left(\frac{1}{(p+a)(p+b)}\right)=\frac{e^{-at}-e^{-bt}}{b-a}$$ for Re$(p+a)\gt 0$ and Re$(p+b)\gt 0$

Therefore $(3)$ becomes $$y=\frac{ae^{-at}-be^{-bt}}{14(a-b)}+\frac{e^{-at}-e^{-bt}}{42(b-a)}\tag{4}$$ finally substitute $a=-1$ and $b=-2$ into $(4)$ to get $$y=\frac{2e^{2t}-e^t}{14}-\frac{e^t-e^{2t}}{42}=\frac{98e^{2t}-56e^t}{588}=\bbox[yellow]{\frac{7e^{2t}-4e^t}{42}}$$

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If you wanted to verify that $$\color{#F80}{\frac{7e^{2t}-4e^t}{42}}$$ is indeed a solution to equation $(1)$: $$y^{\prime\prime} - 3y^{\prime} + 2y = 0$$ then substitute it in: $$y^{\prime}=\frac{14e^{2t}-4e^t}{42}$$ $$y^{\prime\prime}=\frac{28e^{2t}-4e^t}{42}$$

Therefore, $$\require{enclose}\frac{28e^{2t}-4e^t}{42}-3\frac{14e^{2t}-4e^t}{42}+2\frac{7e^{2t}-4e^t}{42}=\frac{1}{42}\left(\enclose{updiagonalstrike}{\color{#180}{28e^{2t}}}\enclose{updiagonalstrike}{\color{blue} {-4e^t}}\enclose{updiagonalstrike}{\color{#180}{-42e^{2t}}}\enclose{updiagonalstrike}{\color{blue}{+12e^t}}\enclose{updiagonalstrike}{\color{#180}{+14e^{2t}}}\enclose{updiagonalstrike}{\color{blue}{-8e^t}}\right)=\frac{1}{42}\left(0\right)=0$$ as required, so $$\bbox[yellow]{y=\frac{7e^{2t}-4e^t}{42}}$$ is a solution to equation $(1)$.