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# Tutor profile: Gyanswarup N.

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Gyanswarup N.
Ph.D. student at Indian Institute of Technology, Indore
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## Questions

### Subject:Linear Algebra

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Question:

Solve the initial value problem $$y'' + 3xy'+3y = 0, y(0) = 1, y'(0) = 0,$$ as a power series in power of $$x$$.

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Gyanswarup N.

For power series solution we assume \begin{align} y & = \sum_{n=0}^{\infty}a_nx^n \end{align} Hence, differentiating with respect to x we get, \begin{align} y' & = \sum_{n=1}^{\infty} na_nx^{n-1}\\ \implies y'' & = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} \end{align} Now we are given the initial value problem, \begin{align*} y'' + 3xy + 3y &= 0 \end{align*} Now putting the values of $$y, y'$$ and $$y''$$ we get, \begin{eqnarray*} \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}+ 3x\sum_{n=1}^{\infty} na_nx^{n-1}+3\sum_{n=0}^{\infty}a_nx^n &=& 0\\ \implies \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+ 3\sum_{n=1}^{\infty} na_nx^{n}+3\sum_{n=0}^{\infty}a_nx^n &=& 0\\ \implies 2a_2+ \sum_{n=1}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+ 3\sum_{n=1}^{\infty} na_nx^{n}+3a_0+3\sum_{n=1}^{\infty}a_nx^n &=& 0\\ \implies 2a_2+3a_0+ \sum_{n=1}^{\infty}\left((n+2)(n+1)a_{n+2}+3na_n +3a_n\right)x^n & = & 0 \\ \implies 2a_2+3a_0+ \sum_{n=1}^{\infty}\left((n+2)(n+1)a_{n+2}+3(n+1)a_n\right)x^n & = & 0\\ \end{eqnarray*} Since each co-efficient of $$x^n$$ must equal to $$0$$, then we have \begin{eqnarray} 2a_2+3a_0 & = & 0\\ (n+1)(n+2)a_{n+2}+3(n+1)a_n & = & 0\\\nonumber \implies a_{n+2} = -\frac{3}{n+2}a_n \end{eqnarray} Again we have the boundary condition $$y(0) = 1$$, and $$y'(0) = 0.$$ Hence, putting the boundary condition we get $$a_0 =1$$ and putting boundary condition on we get, $$a_1= 0.$$ Now using the recurrance relation in we get $$a_{2n} = \frac{(-3)^n}{2^n n!}a_0$$ and $$a_{2n+1}= \frac{(-3)^n}{(2n+1)(2n-1)\cdots3\cdot 1}a_1$$. Therefore the solution $$y = \sum_{n=0}^{\infty}(-1)^n\frac{3^n}{2^n n!}x^{2n}.$$

### Subject:Set Theory

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Question:

Let $$f: \mathbb{R} \mapsto \mathbb{R}$$ be defined by $$f(x) = 3x+2, x \in \mathbb{R}$$. Examine if $$f$$ is (i) injective, (ii) surjective.

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Gyanswarup N.

(i) Let us take two distinct elements $$x_1, x_2 \in \mathbb{R},$$ the domain of $$f$$. $$f(x_1) = 3x_1+2, f(x_2) = 3x_2 + 2.$$ $$f(x_1)-f(x_2) = 3(x_1-x_2) \neq 0, since x_1 \neq x_2.$$ Since $$x_1 \neq x_2 \implies f(x_1) \neq f(x_2),$$ $$f$$ is injective. (ii) Let us take arbitrary element $$y$$ in the set $$\mathbb{R}$$, the co-domain of $$f$$; and let us examine if $$y$$ has a pre-image $$x$$ in the domain of $$f$$. Then $$f(x) = y$$ and therefore $$3x+2 = y$$ or $$x= \frac{y-2}{3}$$. Since $$y \in \mathbb{R}, \frac{y-2}{3} \in \mathbb{R}.$$ Therefore $$y$$ has a pre-image $$\frac{y-2}{3}$$ in the domain of $$f$$. Since $$y$$ is arbitrary, each element in the co-domain of $$f$$ has a pre-image under $$f$$. Therefore $$f$$ is surjective.

### Subject:Differential Equations

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Question:

Find the solution of the system of equations \begin{eqnarray} x+2y+z &=&1\\ 3x+y+2z & = & 3\\ x+7y+2z &= & 1 \end{eqnarray} in integers.

Inactive
Gyanswarup N.

The above system of equations can be written as $$\begin{bmatrix} 1 & 2 & 1\\ 3 & 1 & 2\\ 1 & 7 & 2 \end{bmatrix}\begin{bmatrix} x \\ y\\ z \end{bmatrix} =\begin{bmatrix} 1 \\ 3\\ 1 \end{bmatrix}.$$ This is a non-homogeneous system. The augmented matrix of the system is $$\bar{A} :=\begin{bmatrix} 1 & 2 & 1 & 1\\ 3 & 1 & 2 & 3\\ 1 & 7 & 2 & 1 \end{bmatrix}.$$ Let us apply elementary row operations on $$\bar{A}$$ to reduce it to a row reduced echelon matrix. $$\bar{A}\underset{\overset{R_2-3R_1}{\longrightarrow}}{\overset{R_3-R_1}{\longrightarrow}} \begin{bmatrix} 1 & 2 & 1 & 1\\ 0 & -5 & -1 & 0\\ 0 & 5 & 1 & 0 \end{bmatrix}{\overset{-\frac{1}{5}R_2}{\longrightarrow}} \begin{bmatrix} 1 & 2 & 1 & 1\\ 0 & 1 & \frac{1}{5} & 0\\ 0 & 5 & 1 & 0 \end{bmatrix}\underset{\overset{R_1-2R_2}{\longrightarrow}}{\overset{R_3-5R_2}{\longrightarrow}}\begin{bmatrix} 1 & 0 & \frac{3}{5} & 1\\ 0 & 1 & \frac{1}{5} & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}.$$ Hence the system is equivalent to \begin{eqnarray} x+\frac{3}{5}z & = & 1\\ y+ \frac{1}{5}z & = & 0. \end{eqnarray} Choosing $$z=c,$$ the solution is $$(1-\frac{3}{5}c, -\frac{1}{5}c, c)$$, where $$c\in \mathbb{R}.$$\\ Since the solutions are to be in integers, $$c=5k$$where $$k$$ is an arbitrary integer. Hence the solution is $$(1-3k, -k, 5k)= (1, 0, 0) + k(-3,-1,5), k$$ being an integer.

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