To solve the second-order linear differential equation

$y'' + 4y' + 5y = 0,$

we use the characteristic equation method.

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Step 1: Write the characteristic equation

The characteristic equation corresponding to the given differential equation is:

$r^2 + 4r + 5 = 0.$

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Step 2: Solve the quadratic equation

Solve $r^2 + 4r + 5 = 0$ using the quadratic formula:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

Here, $a = 1$, $b = 4$, and $c = 5$. Substituting these values:

$r = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}.$

Since the discriminant is negative ($-4$), the roots are complex:

$r = \frac{-4 \pm 2i}{2} = -2 \pm i.$

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Step 3: Write the general solution

For complex roots of the form $r = \alpha \pm \beta i$, the general solution is:

$y(t) = e^{\alpha t} \left(C_1 \cos(\beta t) + C_2 \sin(\beta t)\right),$

where:

• $\alpha = -2$,
• $\beta = 1$.

Thus, the general solution is:

$y(t) = e^{-2t} \left(C_1 \cos(t) + C_2 \sin(t)\right),$

where $C_1$ and $C_2$ are arbitrary constants.

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Final Answer:

$y(t) = e^{-2t} \left(C_1 \cos(t) + C_2 \sin(t)\right).$

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Would you like to see a verification of the solution or more details about any step?

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5 Related Questions:

1. How do you solve second-order differential equations with repeated real roots?
2. What is the physical interpretation of a complex root in differential equations?
3. How can you find particular solutions if initial conditions are provided?
4. What is the significance of the exponential term $e^{-2t}$ in the solution?
5. How would the solution change if the coefficients of $y'$ and $y$ were different?

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Tip:

For equations with complex roots, always write the general solution in terms of exponential decay/growth and trigonometric functions for clarity.