Question

determine the response of the damped vibrating system given by the below mentioned differential equation using laplace transform

d2y/dt2+3dy/dt+2y=2(t^2+t+1) ;y(0)=2 ,y'(0)=0

Answer 1

Answer on Question #42602– Math – Differential Calculus | Equation

Question:

Using laplace transforms solve the following ordinary differential equation

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

Solution:

Let $Y(s) = L[y(t)](s)$ . 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

$$L [ y ^ {\prime \prime} + 3 y ^ {\prime} + 2 y ] (s) = L [ 2 (t ^ {2} + t + 1) ] (s)$$

If we look at the left-hand side, we have

$$L [ y ^ {\prime \prime} + 3 y ^ {\prime} + 2 y ] (s) = L [ y ^ {\prime \prime} ] (s) + 3 L [ y ^ {\prime} ] (s) + 2 L [ y ] (s)$$

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

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

Hence, we have

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

If we look at the right-hand side, we have

$$L [ 2 (t ^ {2} + t + 1) ] (s) = 2 L [ t ^ {2} ] (s) + 2 L [ t ] (s) + 2 L [ 1 ] (s)$$

Now use the formulas for the $L[t^2]$ , $L[t]$ and $L[1]$ :

$$L [ t ^ {2} ] = \frac {2}{s ^ {3}}, \qquad L [ t ] = \frac {1}{s ^ {2}}, \quad L [ 1 ] = \frac {1}{s}.$$

The Laplace-transformed differential equation is

$$s ^ {2} Y (s) - s - 2 + 3 (s Y (s) - 1) + 2 Y (s) = \frac {4}{s ^ {3}} + \frac {2}{s ^ {2}} + \frac {2}{s}$$ $$(s ^ {2} + 3 s + 2) Y (s) = s + 5 + \frac {4}{s ^ {3}} + \frac {2}{s ^ {2}} + \frac {2}{s}$$

Or

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

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

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

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

$$Y(s) = \frac{2}{s^3} - \frac{2}{s^2} - \frac{2}{s + 2} + \frac{3}{s} \quad (for\ s \neq -1)$$

Recall the inverse transforms:

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

Using linearity of the inverse transform, we have

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

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

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