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Chapter 5: Problem 1
\(2 \times 4\)
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Substituting \(y=\sum_{n=0}^{\infty} c_{n} x^{n}\) into the differential equation we have $$y^{\prime \prime}+e^{x} y^{\prime}-y=\sum_{n=2}^{\infty} n(n-1) c_{n} x^{n-2}$$ $$\begin{array}{l} \quad+\left(1+x+\frac{1}{2} x^{2}+\frac{1}{6} x^{3}+\cdots\right)\left(c_{1}+2 c_{2} x+3 c_{3} x^{2}+4 c_{4} x^{3}+\cdots\right)-\sum_{n=0}^{\infty} c_{n} x^{n} \\ =\left[2 c_{2}+6 c_{3} x+12 c_{4} x^{2}+20 c_{5} x^{3}+\cdots\right] \\ \quad+\left[c_{1}+\left(2 c_{2}+c_{1}\right) x+\left(3 c_{3}+2 c_{2}+\frac{1}{2} c_{1}\right) x^{2}+\cdots\right]-\left[c_{0}+c_{1} x+c_{2} x^{2}+\cdots\right] \\ =\left(2 c_{2}+c_{1}-c_{0}\right)+\left(6 c_{3}+2 c_{2}\right) x+\left(12 c_{4}+3 c_{3}+c_{2}+\frac{1}{2} c_{1}\right) x^{2}+\cdots=0 \end{array}$$ Thus $$\begin{array}{c} 2 c_{2}+c_{1}-c_{0}=0 \\ 6 c_{3}+2 c_{2}=0 \\ 12 c_{4}+3 c_{3}+c_{2}+\frac{1}{2} c_{1}=0 \end{array}$$ and $$\begin{array}{l} c_{2}=\frac{1}{2} c_{0}-\frac{1}{2} c_{1} \\ c_{3}=-\frac{1}{3} c_{2} \\ c_{4}=-\frac{1}{4} c_{3}+\frac{1}{12} c_{2}-\frac{1}{24} c_{1} \end{array}$$ Choosing \(c_{0}=1\) and \(c_{1}=0\) we find $$c_{2}=\frac{1}{2}, \quad c_{3}=-\frac{1}{6}, \quad c_{4}=0$$ and so on. For \(c_{0}=0\) and \(c_{1}=1\) we obtain $$c_{2}=-\frac{1}{2}, \quad c_{3}=\frac{1}{6}, \quad c_{4}=-\frac{1}{24}$$ and so on. Thus, two solutions are $$y_{1}=1+\frac{1}{2} x^{2}-\frac{1}{6} x^{3}+\cdots \quad \text { and } \quad y_{2}=x-\frac{1}{2} x^{2}+\frac{1}{6} x^{3}-\frac{1}{24} x^{4}+\cdots$$.
The singular points of \(\left(x^{2}-25\right) y^{\prime \prime}+2 x y^{\prime}+y=0\) are -5 and \(5 .\) The distance from 0 to either of these points is \(5 .\) The distance from 1 to the closest of these points is 4.
$$\sin x \cos x=\left(x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}+\cdots\right)\left(1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+\cdots\right)=x-\frac{2 x^{3}}{3}+\frac{2 x^{5}}{15}-\frac{4 x^{7}}{315}+\cdots$$
\(3 \times 2\)
\(\left(\begin{array}{ccc}0 & -\partial / \partial z & \partial / \partial y \\\ \partial / \partial z & 0 & -\partial / \partial x \\ -\partial / \partial y & \partial / \partial x & 0\end{array}\right)\left(\begin{array}{c}P \\ Q \\\ R\end{array}\right)=\left(\begin{array}{r}-\partial Q / \partial z+\partial R / \partial y \\ \partial P / \partial z-\partial R / \partial x \\ -\partial P / \partial y+\partial Q / \partial x\end{array}\right)=\operatorname{curl} \mathbf{F}\)
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