For any real number a the differential equation is given $y^{\prime \prime}(t)+a y^{\prime}(t)+9 y(t)=0$.
a) Determine for $\mathrm{a}=10$ the complete solution to the differential equation. b) Determine whether there are values of the number a for which $y(t)=2 t e^{\wedge}-3 t$ is a solution to the differential equation, and if so, state these values of the number a.

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∻Solution by Steps∻ ‖ step 1 ⋮ Given the differential equation $y''(t) + a y'(t) + 9 y(t) = 0$, we need to determine the complete solution for $a = 10$. ‖ ‖ step 2 ⋮ Substitute $a = 10$ into the differential equation: $y''(t) + 10 y'(t) + 9 y(t) = 0$. ‖ ‖ step 3 ⋮ The characteristic equation for this differential equation is $r^2 + 10r + 9 = 0$. ‖ ‖ step 4 ⋮ Solve the characteristic equation: $r = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{-10 \pm \sqrt{100 - 36}}{2} = \frac{-10 \pm \sqrt{64}}{2} = \frac{-10 \pm 8}{2}$. ‖ ‖ step 5 ⋮ This gives us the roots $r_1 = -1$ and $r_2 = -9$. ‖ ‖ step 6 ⋮ The general solution to the differential equation is $y(t) = C_1 e^{-t} + C_2 e^{-9t}$. ‖ ‖ step 7 ⋮ To determine if $y(t) = 2t e^{-3t}$ is a solution for some value of $a$, substitute $y(t) = 2t e^{-3t}$ into the differential equation $y''(t) + a y'(t) + 9 y(t) = 0$. ‖ ‖ step 8 ⋮ Compute the first derivative: $y'(t) = 2 e^{-3t} - 6t e^{-3t}$. ‖ ‖ step 9 ⋮ Compute the second derivative: $y''(t) = -6 e^{-3t} + 18t e^{-3t}$. ‖ ‖ step 10 ⋮ Substitute $y(t)$, $y'(t)$, and $y''(t)$ into the differential equation: $(-6 e^{-3t} + 18t e^{-3t}) + a (2 e^{-3t} - 6t e^{-3t}) + 9 (2t e^{-3t}) = 0$. ‖ ‖ step 11 ⋮ Simplify the equation: $-6 e^{-3t} + 18t e^{-3t} + 2a e^{-3t} - 6a t e^{-3t} + 18t e^{-3t} = 0$. ‖ ‖ step 12 ⋮ Combine like terms: $(-6 + 2a) e^{-3t} + (18 - 6a + 18) t e^{-3t} = 0$. ‖ ‖ step 13 ⋮ For the equation to hold for all $t$, the coefficients of $e^{-3t}$ and $t e^{-3t}$ must be zero: $-6 + 2a = 0$ and $18 - 6a + 18 = 0$. ‖ ‖ step 14 ⋮ Solve for $a$: $2a = 6 \Rightarrow a = 3$. ‖ ‖ step 15 ⋮ Therefore, $y(t) = 2t e^{-3t}$ is a solution to the differential equation for $a = 3$. ‖ ∻Answer∻ ⚹ The complete solution for $a = 10$ is $y(t) = C_1 e^{-t} + C_2 e^{-9t}$. The function $y(t) = 2t e^{-3t}$ is a solution to the differential equation for $a = 3$. ⚹ ∻Key Concept∻ ⚹Solving differential equations using characteristic equations and verifying solutions.⚹ ∻Explanation∻ ⚹The characteristic equation helps find the general solution to a linear differential equation. Verifying a specific solution involves substituting it back into the original equation and solving for the parameter.⚹◊What is the general solution to the differential equation $y''(t) + 10y'(t) + 9y(t) = 0$?⍭ Generate me a similar question◊

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