Mat 275 Online Activity: The Equation Y = 6y 0

Mat 275 Online Activity 4the Equation 𝑦 𝑦 6𝑦 0 Has The Genera

Mat 275 Online Activity 4 the Equation 𝑦 𝑦 6𝑦 0 Has The Genera

MAT 275 Online Activity 4 The equation ð‘¦â€²â€² − ð‘¦â€² − 6𑦠= 0 has the general solution ð‘¦(ð‘¡) = ð´ð‘’3ð‘¡ + ðµð‘’−2ð‘¡. The exponential function has only positive values. Therefore, if ð´ and ðµ are both positive, then ð‘¦(ð‘¡) is positive for all real numbers ð‘¡. Likewise, if ð´ and ðµ are both negative, then ð‘¦(ð‘¡) is negative for all ð‘¡. Things get more interesting when ð´ and ðµ have opposite signs.

For example, the following is a plot of the particular solution ð‘¦(ð‘¡) = −2ð‘’3ð‘¡ + ð‘’−2ð‘¡: Observe that the solution is negative for all 𑡠≥ 0. (We will only consider behavior of solutions on that interval in this activity.) Another particular solution, ð‘¦(ð‘¡) = −ð‘’3ð‘¡ + 2ð‘’−2ð‘¡ starts out initially positive, but then switches to negative values, where it remains. The plot illustrates this: 1. Explain, using, the algebraic and limit, laws of the exponential function, why, for any negative ð´ and positive ðµ, lim ð‘¡â†’∞ ð‘¦(ð‘¡) = −∞. 2. For each given negative number ð´, there is an interval of positive numbers ðµ so that the corresponding ð‘¦(ð‘¡) will have a zero for some positive ð‘¡. Find that interval for ðµ, in terms of ð´, and the ð‘¡ value of the zero in terms of ð´ and ðµ. 3. Can ð‘¦(ð‘¡) have more than one positive zero for a negative ð´ and positive ðµ?

Paper For Above instruction

The behavior of solutions to second-order linear differential equations, particularly those resembling the form ð‘¦â€²â€² − ð‘¦â€² − 6𑦠= 0, is fundamental in understanding various physical phenomena, including mechanical vibrations and electrical circuits. When examining the general solution ð‘¦(ð‘¡) = ð´ð‘’3ð‘¡ + ðµð‘’−2ð‘¡, we see that the exponential functions involved play a critical role in the qualitative behavior of solutions, especially when considering the signs and limits of the parameters ð´ and ðµ.

Part 1: Limit Behavior for Negative ð´ and Positive ðµ

The exponential function e^{kð‘¡} is always positive for real ð‘¡, regardless of whether ð‘’ is positive or negative. Specifically, for any real number ð‘’, the limit of e^{kð‘¡} as ð‘¡ approaches infinity depends on the sign of ð‘’. If ð‘’ > 0, then lim ð‘¡→∞ e^{kð‘¡}} = ∞; if ð‘’

For negative ð´ ( 0), the exponentials behave as follows: e^{3ð‘¡} with ð‘’ > 0 tends to infinity exponentially as ð‘¡ → ∞, but with ð‘’ 0), the product ð´ e^{3ð‘¡} will tend toward (−)∞ as ð‘¡ → ∞ when ð‘’ 0, this exponential grows without bound, and since ð’ 0, tends to zero, and with ð’ > 0, it tends to zero, but when ð‘’ > 0, and ð‘’ > 0, the exponential diminishes but remains positive. Now, considering the entire sum, for all ð‘¡ greater than some large number, the solution ð‘¦(ð‘¡) tends to (−)∞, regardless of the relative sizes of ð‘’ and ð‘’.

Thus, for negative ð‘’ and positive ð‘’, lim ð‘¡→∞ ð‘¦(ð‘¡) = −∞. This conclusion stems from the dominance of the exponential terms' signs and their limits dictated by the exponential laws. The negative coefficient associated with the ð‘’3ð‘¡} term dictates the solution's ultimate tendency toward negative infinity as ð‘¡ becomes large, confirming the asymptotic unboundedness in the negative direction.

Part 2: Intervals of 𑵠for Zeros of the Solution

Next, we analyze conditions under which ð‘¦(ð‘¡) = 0 for some positive ð‘¡. The zeros of ð‘¦(ð‘¡) are solutions to ð´ e^{3ð‘¡} + ðµ e^{−2ð‘¡}} = 0. Rearranging gives:

ð´ e^{3ð‘¡}} = − ðµ e^{−2ð‘¡}}. Since this equality involves two exponential terms, both positive when considering e^{að‘¡}}, the right side must be negative for the equality to hold, meaning ð´ and ðµ must have opposite signs, which aligns with the initial conditions for this part of the analysis.

Dividing both sides by e^{−2ð‘¡}}, which is positive, yields:

ð´ e^{5ð‘¡}} = - ðµ.

Solving for ð‘¡ gives:

e^{5ð‘¡}} = - ðµ / ð´.

Since the exponential function is always positive, the right side must be positive, which implies:

• ðµ / ð´}

Thus, the zero occurs at:

ð‘¡ = (1/5) * ln (- ðµ / ð´}).

This gives the value of ð‘¡ of the zero in terms of ð‘´ and ð‘µ. The interval of 𑵠for a given negative ð‘´ such that the zero exists for some ð‘¡ > 0 is all 𑵠satisfying:

- ð‘´ e^{5ð‘¡}} = ðµ, with ð‘¡ > 0, which requires ðµ in the interval (- ∞, 0) if ð‘´ 0, the roles are reversed in terms of signs, but primarily for the zero to be positive, the same conditions apply.

Part 3: Multiple Positive Zeros?

Finally, whether ð‘¦(ð‘¡) can have more than one positive zero given negative ð‘´ and positive 𑵠depends on the nature of solutions of the equation ð‘´ e^{3ð‘¡} + 𑵠e^{−2ð‘¡}} = 0. Since this is a sum of exponential functions with different exponents, it is possible for the function to cross zero multiple times, especially if it oscillates or has local extrema. Specifically, because the sum involves an exponential growth and decay, the graph of ð‘¦(ð‘¡) can intersect the zero line at multiple points, provided the parameters ð‘´ and 𑵠are such that the function exhibits multiple sign changes.

In conclusion, the exponential form of the solution indicates that for negative ð‘´ and positive ð‘µ, the solution diverges to negative infinity as ð‘¡ approaches infinity, consistent with the dominant behavior of exponential terms. Moreover, the zeros of the function depend logarithmically on these parameters, and multiple zeros are possible when the exponential sum oscillates due to the interplay of the growth and decay rates dictated by ð‘’ and ð‘’.

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