To solve the second-order initial value problem using the Laplace transform, let’s go through the process step-by-step.

Problem Statement

Given: $y'' + 7y' + 10y = 0$ with initial conditions: $y(0) = 2, \quad y'(0) = 1$

Step 1: Take the Laplace Transform of Both Sides

The Laplace transform of $y''$ and $y'$ are: $\mathcal{L}(y'') = s^2 Y(s) - s y(0) - y'(0)$ $\mathcal{L}(y') = s Y(s) - y(0)$ where $Y(s) = \mathcal{L}(y)$.

Applying the Laplace transform to both sides of the differential equation: $\mathcal{L}(y'') + 7\mathcal{L}(y') + 10\mathcal{L}(y) = 0$

Step 2: Substitute the Transforms and Initial Conditions

Substitute $\mathcal{L}(y'')$, $\mathcal{L}(y')$, and $\mathcal{L}(y)$ using the initial conditions $y(0) = 2$ and $y'(0) = 1$:

$(s^2 Y(s) - s \cdot 2 - 1) + 7(s Y(s) - 2) + 10Y(s) = 0$

Expanding and simplifying: $s^2 Y(s) - 2s - 1 + 7s Y(s) - 14 + 10 Y(s) = 0$

Combine like terms involving $Y(s)$: $(s^2 + 7s + 10)Y(s) = 2s + 15$

Step 3: Solve for $Y(s)$

Now, isolate $Y(s)$: $Y(s) = \frac{2s + 15}{s^2 + 7s + 10}$

Step 4: Factor the Denominator and Decompose into Partial Fractions

The denominator factors as: $s^2 + 7s + 10 = (s + 5)(s + 2)$

Thus, $Y(s) = \frac{2s + 15}{(s + 5)(s + 2)}$

We can express this in partial fraction form: $Y(s) = \frac{A}{s + 5} + \frac{B}{s + 2}$

Multiply both sides by $(s + 5)(s + 2)$ to find $A$ and $B$: $2s + 15 = A(s + 2) + B(s + 5)$

Expanding: $2s + 15 = As + 2A + Bs + 5B$

Group terms: $2s + 15 = (A + B)s + (2A + 5B)$

Equating coefficients:

1. $A + B = 2$
2. $2A + 5B = 15$

Solve this system of equations for $A$ and $B$:

• From the first equation, $B = 2 - A$.
• Substitute into the second equation: $2A + 5(2 - A) = 15$ $2A + 10 - 5A = 15$ $-3A = 5$ $A = -\frac{5}{3}$ Substituting $A = -\frac{5}{3}$ into $B = 2 - A$: $B = 2 + \frac{5}{3} = \frac{6 + 5}{3} = \frac{11}{3}$

Thus, $Y(s) = \frac{-5/3}{s + 5} + \frac{11/3}{s + 2}$

Step 5: Take the Inverse Laplace Transform

Now, take the inverse Laplace transform of each term separately: $y(t) = -\frac{5}{3}e^{-5t} + \frac{11}{3}e^{-2t}$

Final Answer

$y(t) = -\frac{5}{3}e^{-5t} + \frac{11}{3}e^{-2t}$

Would you like a detailed verification of each step or have any questions?

Here are some related questions for further exploration:

1. How would the solution change if different initial conditions were given?
2. Can you solve a similar second-order differential equation with complex roots?
3. What are the steps for solving non-homogeneous differential equations with the Laplace transform?
4. How does the Laplace transform handle derivatives of non-zero initial values?
5. What is the significance of the characteristic roots in determining the behavior of the solution?

Tip: Always check the characteristic roots to anticipate the nature of the solution (real, complex, or repeated roots).