Question

using laplace transforms solve the following ordinary differential equation
d3y/dt3+2d2y/dt2-dy/dt-2y=o ,y(0)=1 ,y'(0)=2 ,y"(0)=2

Answer 1

Answer on Question #42499– Math – Other

Question:

Using laplace transforms solve the following ordinary differential equation

$$\frac {\mathrm {d} ^ {3} y}{\mathrm {d t} ^ {3}} + 2 \frac {\mathrm {d} ^ {2} y}{\mathrm {d t} ^ {2}} - \frac {\mathrm {d} y}{\mathrm {d t}} - 2 y = 0, y (0) = 1, y ^ {\prime} (0) = 2, y ^ {\prime \prime} (0) = 2$$

Solution:

Let $Y(s) = L\gamma(t)$ . Instead of solving directly for $y(t)$ , we derive a new equation for $Y(s)$ . Once we find $Y(s)$ , we inverse transform to determine $y(t)$ .

The first step is to take the Laplace transform of both sides of the original differential equation. We have

$$L \left[ y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} - y ^ {\prime} - 2 y \right] (s) = L [ 0 ] (s)$$

Obviously, the Laplace transform of the function 0 is 0. If we look at the left-hand side, we have

$$L \left[ y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} - y ^ {\prime} - 2 y \right] (s) = L \left[ y ^ {\prime \prime \prime} \right] (s) + 2 L \left[ y ^ {\prime \prime} \right] (s) - L \left[ y ^ {\prime} \right] (s) - 2 L \left[ y \right] (s)$$

Now use the formulas for the $L[y''']$ , $L[y'']$ and $L[y']$ :

$$L \left[ y ^ {\prime} \right] = s L [ y ] - y (0) = s Y (s) - 1;$$ $$L \left[ y ^ {\prime \prime} \right] = s ^ {2} L [ y ] - s y (0) - y ^ {\prime} (0) = s ^ {2} Y (s) - s - 2;$$ $$L \left[ y ^ {\prime \prime \prime} \right] = s ^ {3} L [ y ] - s ^ {2} y (0) - s y ^ {\prime} (0) - y ^ {\prime \prime} (0) = s ^ {3} Y (s) - s ^ {2} - 2 s - 2;$$

Hence, we have

$$L \left[ y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} - y ^ {\prime} - 2 y \right] (s) = s ^ {3} Y (s) - s ^ {2} - 2 s - 2 + 2 \left(s ^ {2} Y (s) - s - 2\right) - \left(s Y (s) - 1\right)$$

The Laplace-transformed differential equation is

$$(s ^ {3} + 2 s ^ {2} - s - 2) Y (s) - s ^ {2} - 2 s - 2 - 2 s - 4 + 1 = 0$$

Or

$$(s ^ {3} + 2 s ^ {2} - s - 2) Y (s) - s ^ {2} - 4 s - 5 = 0.$$

This is a linear algebraic equation for $Y(s)!$ . We have converted a differential equation into a algebraic equation! Solving for $Y(s)$ , we have

$$Y (s) = \frac {s ^ {2} + 4 s + 5}{s ^ {3} + 2 s ^ {2} - s - 2}$$

We can simplify this expression using the method of partial fractions:

$$Y(s) = -\frac{1}{s + 1} + \frac{1}{3(s + 2)} + \frac{5}{3(s - 1)}$$

Recall the inverse transforms:

$$L^{-1}\left[\frac{1}{s + 1}\right](t) = e^{-t}, \quad L^{-1}\left[\frac{1}{s + 2}\right](t) = e^{-2t}, \quad L^{-1}\left[\frac{1}{s - 1}\right](t) = e^{t}$$

Using linearity of the inverse transform, we have

$$y(t) = L^{-1}\left[ -\frac{1}{s + 1} + \frac{1}{3(s + 2)} + \frac{5}{3(s - 1)} \right] = -e^{-t} + \frac{1}{3}e^{-2t} + \frac{5}{3}e^{t} = \frac{1}{3}e^{-2t}(5e^{3t} - 3e^{t} + 1);$$

Answer. $y(t) = \frac{1}{3}e^{-2t}(5e^{3t} - 3e^{t} + 1)$;

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