Question

1. There is a large class of differential equations—the so-called linear ones— for which we can find solutions using the Taylor series method discussed in the Lecture. One such differential equation is $$t\frac{d^2x}{dt^2} + \frac{dx}{dt} + tx = 0 \quad ()$$ It is a particular case of the more general Bessel differential equation, and one solution of it is given by the Bessel function $J_0(t)$ that we saw in Chapter 1. Notice that $()$ involves not only the first derivative $\frac{dx}{dt}$ but also the second derivative $\frac{d^2x}{dt^2}$. For this reason, it is said to be a second order differential equation. In this problem we will content ourselves with finding a relationship (specifically, a recurrence relation) on the coefficients of a Taylor series expansion about $t = 0$ of a solution to our equation. Hence consider the Taylor series $$x(t) = \sum_{k=0}^{\infty} c_k t^k$$ Substituting this into $(*)$ will give you two conditions. The first one is $c_1 = 0$. What is the other one? Note: this problem involves some nontrivial manipulation of indices in summation notation. Do not get discouraged if it feels more difficult than other problems: it is! $$c_k = -\frac{c_{k-1}}{(k-1)^2}$$ $$c_k = \frac{c_{k-2}}{k^2}$$ $$c_k = -\frac{c_{k-2}}{(k-2)^2}$$ $$c_k = -\frac{c_{k-1}}{k^2}$$ $$c_k = \frac{c_{k-1}}{k^2}$$ $$c_k = -\frac{c_{k-2}}{k^2}$$

          1. There is a large class of differential equations—the so-called linear ones— for which we can find solutions using the Taylor series method discussed in the Lecture. One such differential equation is
$$t\frac{d^2x}{dt^2} + \frac{dx}{dt} + tx = 0 \quad (*)$$
It is a particular case of the more general Bessel differential equation, and one solution of it is given by the Bessel function $J_0(t)$ that we saw in Chapter 1. Notice that $(*)$ involves not only the first derivative $\frac{dx}{dt}$ but also the second derivative $\frac{d^2x}{dt^2}$. For this reason, it is said to be a second order differential equation.
In this problem we will content ourselves with finding a relationship (specifically, a recurrence relation) on the coefficients of a Taylor series expansion about $t = 0$ of a solution to our equation. Hence consider the Taylor series
$$x(t) = \sum_{k=0}^{\infty} c_k t^k$$
Substituting this into $(*)$ will give you two conditions. The first one is $c_1 = 0$. What is the other one?
Note: this problem involves some nontrivial manipulation of indices in summation notation. Do not get discouraged if it feels more difficult than other problems: it is!
$$c_k = -\frac{c_{k-1}}{(k-1)^2}$$
$$c_k = \frac{c_{k-2}}{k^2}$$
$$c_k = -\frac{c_{k-2}}{(k-2)^2}$$
$$c_k = -\frac{c_{k-1}}{k^2}$$
$$c_k = \frac{c_{k-1}}{k^2}$$
$$c_k = -\frac{c_{k-2}}{k^2}$$

1 there is a large class of differential equationsthe so called linear ones for which we can find solutions using the taylor series method discussed in the lecture one such differential equa 76813

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