I see you've uploaded an image containing solutions for two problems: solving an initial value problem (IVP) and finding the Laplace transform of $t^2 \cos(bt)$. I'll simplify and explain the solutions step by step for better clarity.

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Problem 6: Solve the initial value problem

The differential equation is: $y'' + 7y' + 10y = 0, \quad y(0) = 4, \quad y'(0) = 2.$

Step 1: Solve the characteristic equation The characteristic equation of the differential equation is: $r^2 + 7r + 10 = 0.$ Factoring gives: $(r + 5)(r + 2) = 0 \implies r = -5, \, -2.$ Thus, the general solution is: $y = C_1 e^{-5t} + C_2 e^{-2t}.$

Step 2: Apply initial conditions

1. Using $y(0) = 4$: $C_1 e^{0} + C_2 e^{0} = 4 \implies C_1 + C_2 = 4.$

2. Using $y'(0) = 2$: Differentiate $y$: $y'(t) = -5C_1 e^{-5t} - 2C_2 e^{-2t}.$ At $t = 0$: $-5C_1 - 2C_2 = 2.$

Now solve the system of equations: $C_1 + C_2 = 4, \quad -5C_1 - 2C_2 = 2.$ Multiply the first equation by 2: $2C_1 + 2C_2 = 8.$ Add to the second equation: $-5C_1 - 2C_2 + 2C_1 + 2C_2 = 2 + 8 \implies -3C_1 = 10 \implies C_1 = -\frac{10}{3}.$ Substitute $C_1 = -\frac{10}{3}$ into $C_1 + C_2 = 4$: $-\frac{10}{3} + C_2 = 4 \implies C_2 = \frac{22}{3}.$

Thus: $C_1 = -\frac{10}{3}, \quad C_2 = \frac{22}{3}.$

The solution is: $y = -\frac{10}{3} e^{-5t} + \frac{22}{3} e^{-2t}.$

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Problem 7: Find the Laplace transform of $t^2 \cos(bt)$

We use the formula: $\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} \mathcal{L}\{f(t)\},$ where $f(t) = \cos(bt)$.

Step 1: Find $\mathcal{L}\{\cos(bt)\}$: $\mathcal{L}\{\cos(bt)\} = \frac{s}{s^2 + b^2}.$

Step 2: Differentiate twice To find $\mathcal{L}\{t^2 \cos(bt)\}$, differentiate $\frac{s}{s^2 + b^2}$ twice with respect to $s$: $\mathcal{L}\{t^2 \cos(bt)\} = \frac{d^2}{ds^2} \left( \frac{s}{s^2 + b^2} \right).$

1. First derivative: $\frac{d}{ds} \left( \frac{s}{s^2 + b^2} \right) = \frac{(s^2 + b^2)(1) - s(2s)}{(s^2 + b^2)^2} = \frac{b^2 - s^2}{(s^2 + b^2)^2}.$

2. Second derivative: Differentiate again: $\frac{d^2}{ds^2} \left( \frac{s}{s^2 + b^2} \right) = \frac{d}{ds} \left( \frac{b^2 - s^2}{(s^2 + b^2)^2} \right).$ Using the quotient rule and simplifying, this becomes: $\frac{d^2}{ds^2} \left( \frac{s}{s^2 + b^2} \right) = \frac{2s(s^2 - 3b^2)}{(s^2 + b^2)^3}.$

Thus: $\mathcal{L}\{t^2 \cos(bt)\} = \frac{2s(s^2 - 3b^2)}{(s^2 + b^2)^3}.$

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Summary of Solutions

1. Problem 6: $y = -\frac{10}{3} e^{-5t} + \frac{22}{3} e^{-2t}.$

2. Problem 7: $\mathcal{L}\{t^2 \cos(bt)\} = \frac{2s(s^2 - 3b^2)}{(s^2 + b^2)^3}.$

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Would you like further details or explanations? Let me know! Here are some related questions to expand on this:

1. How do you find the Laplace transform of other products like $t \sin(bt)$?
2. What are the physical interpretations of these solutions in applied scenarios?
3. Can you solve the IVP using numerical methods or software?
4. What is the general method for solving higher-order linear differential equations?
5. How can Laplace transforms be used to solve initial value problems directly?

Tip: When using Laplace transforms, always simplify the transform before applying differentiation to make calculations easier.