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Q21E

Expert-verifiedFound in: Page 332

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**Solve the given initial value problem **

_{$\begin{array}{l}y\text{'}\text{'}\text{'}-4y\text{'}\text{'}+7y\text{'}-6y=0\\ y\left(0\right)=1\\ y\text{'}\left(0\right)=0\\ y\text{'}\text{'}\left(0\right)=0\end{array}$}

_{The solution is }_{$y\left(t\right)={e}^{2t}-\sqrt{2}{e}^{t}\mathrm{sin}\sqrt{2}t$}

The Sum rule says the derivative of a sum of functions is the sum of their derivatives. The Difference rule says the derivative of a difference of functions is the difference of their derivatives.

We will do the following question on the basis of basic differentiation $\begin{array}{l}{r}^{3}-4{r}^{2}+7r-6=0\\ {r}^{3}-4{r}^{2}+7r-6=(r-2)({r}^{2}-2r+3)=0\\ \\ y\left(t\right)={c}_{1}{e}^{2t}+{c}_{2}{e}^{t}\mathrm{sin}\sqrt{2}t+{c}_{3}{e}^{t}\mathrm{cos}\sqrt{2}t\\ y\text{'}\left(t\right)=2{c}_{1}{e}^{2t}+{c}_{2}{e}^{t}\mathrm{sin}\sqrt{2}t+\sqrt{2}{c}_{2}{e}^{t}\mathrm{cos}\sqrt{2}t+{c}_{3}{e}^{t}\mathrm{cos}\sqrt{2}t-\sqrt{2}{c}_{3}{e}^{t}\mathrm{sin}\sqrt{2}t\\ y\text{'}\text{'}\left(t\right)=4{c}_{1}{e}^{2t}+{c}_{2}{e}^{t}\mathrm{sin}\sqrt{2}t+\sqrt{2}{c}_{2}{e}^{t}\mathrm{cos}\sqrt{2}t+\sqrt{2}{c}_{2}{e}^{t}\mathrm{cos}\sqrt{2}t-\sqrt{2}{c}_{2}{e}^{t}\mathrm{sin}\sqrt{2}t+{c}_{3}{e}^{t}\mathrm{cos}\sqrt{2}t-\sqrt{2}{c}_{3}{e}^{t}\mathrm{sin}\sqrt{2}t-\sqrt{2}{c}_{3}{e}^{t}\mathrm{sin}\sqrt{2}t-\sqrt{2}{c}_{3}{e}^{t}\mathrm{cos}\sqrt{2}t\\ \\ y\left(0\right)=1\\ y\text{'}\left(0\right)=0\\ y\text{'}\text{'}\left(0\right)=0\\ \\ y\left(t\right)={e}^{2t}-\sqrt{2}{e}^{t}\mathrm{sin}\sqrt{2}t\end{array}$

Hence, the final answer is:

_{}

$y\left(t\right)={e}^{2t}-\sqrt{2}{e}^{t}\mathrm{sin}\sqrt{2}t$

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