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2006 SMT/Calculus Problems/Problem 2: Difference between revisions

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\frac{0}{e^{\lambda t}}&=\frac{1}{e^{\lambda t}}(4\lambda^{2}e^{\lambda t}+3\lambda e^{\lambda t}-e^{\lambda t})\\
\frac{0}{e^{\lambda t}}&=\frac{1}{e^{\lambda t}}(4\lambda^{2}e^{\lambda t}+3\lambda e^{\lambda t}-e^{\lambda t})\\
0&=4\lambda^{2}+3\lambda - 1\\
0&=4\lambda^{2}+3\lambda - 1\\
0&=(4\lambda-1)(\lambda+1) \Rightarrow \lambda = \frac{1}{4}, 1\\
0&=(4\lambda-1)(\lambda+1) \Rightarrow \lambda = \frac{1}{4}, -1\\
\end{align*}</cmath>
\end{align*}</cmath>


Therefore, the possible values of <math>\lambda</math> are <math>\boxed{\lambda = \frac{1}{4}, 1}</math>
Therefore, the possible values of <math>\lambda</math> are <math>\boxed{\lambda = \frac{1}{4}, -1}</math>

Latest revision as of 11:37, 15 January 2020

Problem 2

Given the equation $4y''+3y'-y=0$ and its solution $y=e^{\lambda t}$, what are the values of $\lambda$?

Solution

Substituting $y=e^{\lambda t}$ into the differential equation, we can solve for the values of $\lambda$:

\begin{align*} 0&=4y''+3y'-y, \quad y=e^{\lambda t}\\ 0&=4\lambda^{2}e^{\lambda t}+3\lambda e^{\lambda t}-e^{\lambda t} \end{align*}

Dividing through by $e^{\lambda t}$: \begin{align*} \frac{0}{e^{\lambda t}}&=\frac{1}{e^{\lambda t}}(4\lambda^{2}e^{\lambda t}+3\lambda e^{\lambda t}-e^{\lambda t})\\ 0&=4\lambda^{2}+3\lambda - 1\\ 0&=(4\lambda-1)(\lambda+1) \Rightarrow \lambda = \frac{1}{4}, -1\\ \end{align*}

Therefore, the possible values of $\lambda$ are $\boxed{\lambda = \frac{1}{4}, -1}$

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