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let $y(x)$ be a solution of the differential equation $y''=y'+y$ satisfying the initial conditions $y(0)=4$ and $y'(0)=7$ . What is the value of $y^{(5)}(0)$ , the fifth derivative of $y$ evaluated at zero ?

I have found,

$$y(x)=\exp({\frac{x}{2})} [4 \cosh (\sqrt{5}/2)+2\sqrt{5} \sinh (\sqrt{5}/2)]$$

But it is not easy to find $y^{(5)}(0)$ in 4 minutes ( actually , it was asked in an examination where we had to solve $15$ multiple select questions in an hour ).

Is there any other method or way to find its solution ?

Thanks !

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We have $$\eqalign{ y''=y'+y&\quad\Rightarrow\quad y''(0)=y'(0)+y(0)=11\cr y'''=y''+y'&\quad\Rightarrow\quad y'''(0)=y''(0)+y'(0)=18\cr}$$ and so on.

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  • $\begingroup$ thanks ! , it was really very easy .....my bad , I couldn't do it in the exam. ($y^{5}(0)=47$) $\endgroup$ – освящение Jun 1 '18 at 7:10
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For the sake of completing your method, note that: $y''=y'+y, y(0)=4, y'(0)=7$ has a solution: $$k^2-k-1=0 \Rightarrow k_{1,2}=\frac{1\pm\sqrt{5}}{2};\\ y(x)=C_1e^{\frac{1+\sqrt{5}}{2}x}+C_2e^{\frac{1-\sqrt{5}}{2}x};\\ \begin{cases} C_1+C_2=4 \\ C_1\frac{1+\sqrt{5}}{2}+C_2\frac{1-\sqrt{5}}{2}=7\end{cases} \Rightarrow \begin{cases}C_1=2+\sqrt{5} \\ C_2=2-\sqrt{5}\end{cases}\\ y=(2+\sqrt{5})e^{\frac{1+\sqrt{5}}{2}x}+(2-\sqrt{5})e^{\frac{1-\sqrt{5}}{2}x}.$$ Hence: $$\begin{align}y^{(5)}(0)=&(2+\sqrt{5})\left(\frac{1+\sqrt{5}}{2}\right)^5+(2-\sqrt{5})\left(\frac{1-\sqrt{5}}{2}\right)^5=\\ =&\frac{(2+\sqrt{5})(3+\sqrt{5})^2(1+\sqrt{5})+(2-\sqrt{5})(3-\sqrt{5})^2(1-\sqrt{5})}{8}=\\ =&\frac{(7+3\sqrt{5})^2+(7-3\sqrt{5})^2}{4}=\\ &=\frac{49+45}{2}=\\ &=47.\end{align}$$

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  • $\begingroup$ thanks ! for answering ... $\endgroup$ – освящение Jun 1 '18 at 10:10
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    $\begingroup$ You are welcome. Good luck on the exam. $\endgroup$ – farruhota Jun 1 '18 at 10:16

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