Question

Use the Laplace transform to solve the following differential equations:

(D² +3D+2)y(t)=Dx(t) if y(0⁻ )= y​ (0²⁻ )=0 and x(t)=u(t)

Answer 1

To solve the given differential equation using Laplace transform, we apply the transform to both sides of the equation. After finding the Laplace transform of x(t), we substitute it into the equation and solve for Y(s). Finally, taking the inverse Laplace transform of Y(s) gives the solution to the differential equation.

Step-by-step explanation:

To solve the differential equation using Laplace transform, we need to apply the transform to both sides of the equation. Applying the Laplace transform to the left-hand side, we get:

$(s^2+3s+2)Y(s) = DX(s)$

where $Y(s)$ and $X(s)$ are the Laplace transforms of $y(t)$ and $x(t)$, respectively. Next, we need to find the Laplace transform of $x(t)$. Since $x(t)$ is the unit step function $u(t)$, its Laplace transform is:

$X(s) = \frac{1}{s}$

Substituting this into the equation, we have:

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

Solving for $Y(s)$, we get:

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

To find the inverse Laplace transform of $Y(s)$, we need to decompose it into partial fractions:

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

Finally, taking the inverse Laplace transform of each term, we get the solution to the differential equation:

$y(t) = 1-e^{-t}+e^{-2t}$