Using Laplace Transform to solve this ODE

Question

How to solve this ODE, with Laplace Transform:

$$ \begin{cases} 20y'(x)+y(x)+4y''(x)=20\\ y(0)=10\\ 4y'(0)=-2 \end{cases} $$

Thanks in advance.

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My work:

$$20y'(x)+y(x)+4y''(x)=20\Longleftrightarrow$$ $$\mathcal{L}_{t}\left[20y'(x)+y(x)+4y''(x)\right]_{(s)}=\mathcal{L}_{t}\left[20\right]_{(s)}\Longleftrightarrow$$ $$20sy(s)-y(0)+y(s)+4s^2y(s)-sy(0)-y'(0)=\frac{20}{s}$$

Answer 1

You are almost there. Let $Y(s)$ be the Laplace transform of $y(t)$.

The last step should actually read:

$20sY(s) - 20y(0) + Y(s) + 4s^2Y(s) -4sy(0) - 4y'(0) = \frac{20}{s}$

Now lets find an expresion for $Y(s)$:

$Y(s)(4s^2 + 20s +1) = \frac{20}{s} + 20y(0) + 4sy(0) + 4y'(0)$

Simplify: $Y(s) = \frac{20 +198s+40s^2}{s(4s^2 + 20s +1)} = \frac{40s}{4s^2 + 20s +1} + \frac{198}{4s^2 + 20s +1} + \frac{20}{s(4s^2 + 20s +1)}$

Note: $4s^2 + 20s +1 = (2s+5)^2 - 24$. Now using Laplace transform table you can easily transform $Y(s)$ back to $y(t)$ using 19, 20 and 32.

This is a fairly brute force approach to the problem, but never the less it works.