Tag: Differential Equations

Determine the interval of convergence of ∑ xⁿ / n! and show that it satisfies y′ = x + y

by RoRi

April 20, 2016

Consider the function $f(x)$ defined by the power series

$f(x) = \sum_{n=2}^{\infty} \frac{x^n}{n!}.$

Determine the interval of convergence for $f(x)$ and show that it satisfies the differential equation

$y' = x + y.$

(We might notice that this is almost the power series expansion for the exponential function $e^x$ and deduce the interval of convergence and the differential equation from properties of the exponential that we already know. We can do it from scratch just as easily though.)

First, we apply the ratio test

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{n+1}}{(n+1)!} \right) \left( \frac{n!}{x^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x}{n+1} \\[9pt] &= 0. \end{align*}$

Hence, the series converges for all $x$ . Next, we take a derivative

$\begin{align*} y &= \sum_{n=2}^{\infty} \frac{x^n}{n!} \\[9pt] y' &= \sum_{n=2}^{\infty} \frac{nx^{n-1}}{n!} \\[9pt] &= \sum_{n=2}^{\infty} \frac{x^{n-1}}{(n-1)!} \\[9pt] &= \sum_{n=1}^{\infty} \frac{x^n}{n!}. \end{align*}$

Then we have

$\begin{align*} x + y &= x + \sum_{n=2}^{\infty} \frac{x^n}{n!} \\[9pt] &= \sum_{n=1}^{\infty} \frac{x^n}{n!} \\[9pt] &= y'. \end{align*}$

Now, to compute the sum we can solve the given differential equation

$y' = x + y \quad \implies \quad y' - y = x.$

This is a first order linear differential equation of the form $y' + P(x)y = Q(x)$ with $P(x) = -1$ and $Q(x) = x$ . We also know that $f(0) = 0$ ; therefore, this equation has a unique solution satisfying the given initial condition which is given by

$y = be^{-A(x)} + e^{-A(x)} \int_a^x Q(t) e^{A(t)} \, dt.$

Where $a = b = 0$ and

$A(x) = \int_0^x P(t) \, dt = - \int_0^x dt = -x.$

Therefore, we have

$\begin{align*} f(x) &= e^{x} \left( \int_0^x t e^{-t} \, dt \\[9pt] &= e^x \left( -te^{-t} \Bigr \rvert_0^x + \int_0^x e^{-t} \, dt \right) \\[9pt] &= e^x \left( -xe^{-x} - e^{-x} + 1 \right) \\[9pt] &= e^x - x - 1. \end{align*}$

Determine the interval of convergence of ∑ x²ⁿ / n! and show it satisfies a given differential equation

by RoRi

April 20, 2016

Consider the function $f(x)$ defined by the power series

$f(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}.$

Determine the interval of convergence for $f(x)$ and show that it satisfies the differential equation

$y' = 2xy.$

First, to determine the interval of convergence we use the ratio test,

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{2n+2}}{(n+1)!} \right) \left( \frac{n!}{x^{2n}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x^2}{n+1} \\[9pt] &= 0. \end{align*}$

Hence, the series converges for all $x$ .

To show that it satisfies the given differential equation, we first take the derivative,

$\begin{align*} && f(x) &= \sum_{n=0}^{\infty} \frac{x^{2n}}{n!} \\[9pt] \implies && f'(x) &= \sum_{n=1}^{\infty} \frac{2n x^{2n-1}}{n!} \\[9pt] &&&= 2 \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(n-1)!}. \end{align*}$

Then, it satisfies the given differential equation since

$\begin{align*} 2xy &= 2x \left( \sum_{n=0}^{\infty} \frac{x^{2n}}{n!} \right)\\[9pt] &= 2 \left( \sum_{n=0}^{\infty} \frac{x^{2n+1}}{n!} \right)\\[9pt] &= 2 \sum_{n=1}^{\infty} \frac{x^{2(n-1)+1}}{(n-1)!} \\[9pt] &= 2 \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(n-1)!} \\[9pt] &= y'. \end{align*}$

Then, since the given differential equation is a first-order linear differential equation of the form

$y' + P(x)y = 0$

with $P(x) = -2x$ we know that the solutions are uniquely determined by the formula

$f(x) = be^{-A(x)}, \qquad \text{where} \qquad A(x) = \int_a^x P(t) \, dt.$

Since we have the initial condition $f(0) = 1$ (by plugging in $x = 0$ to the power series expansion for $f(x)$ ) we have $a= b= 0$ and the unique solution of this differential equation is

$f(x) = e^{-A(x)} = \exp \left( - \int_0^x (-2t) \, dt \right) = e^{x^2}.$

Determine the interval of convergence of a given power series and show that it satisfies a given differential equation

by RoRi

April 20, 2016

Consider the function $f(x)$ defined by the power series

$f(x) = 1 + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n}.$

Determine the interval of convergence for $f(x)$ and show that it satisfies the differential equation

$y'' = x^a y + b.$

Find $a$ and $b$ .

First, to determine the interval of convergence for $f(x)$ we use the ratio test,

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{1 \cdot 4 \cdot 7 \cdots (3n+1) \cdot x^{3n+3}}{(3n+3)!} \right) \left( \frac{(3n)!}{1 \cdot 4 \cdot 7 \cdots (3n-2) \cdot x^{3n}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{(3n+1)x^3}{(3n+1)(3n+2)(3n+3)} \\[9pt] &= \lim_{n \to \infty} \frac{x^3}{(3n+2)(3n+3)} \\[9pt] &= 0. \end{align*}$

Therefore, the series converges for all $x$ . Next, to show that it satisfies the given differential equation, we take the first two derivatives,

$\begin{align*} y &= 1 + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n} \\[9pt] y' &= \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)\cdot (3n)}{(3n)!} x^{3n-1} = \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-1)!} x^{3n-1} \\[9pt] y'' &= \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)\cdot (3n-1)}{(3n-1)!} x^{3n-2} = \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!}x^{3n-2}. \end{align*}$

Then, we have the differential equation $y'' = x^a y + b$ ,

$\begin{align*} && y'' &= x^a y + b \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!}x^{3n-2} &= x^a\left( 1 + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n} \right) + b \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!} x^{3n-2} &= b + x^a + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n+a}. \end{align*}$

But, for this equation to hold we must have $b = 0$ (since there is no constant term in $y''$ on the left) and we must also have $a = 1$ since there the coefficient of $x^1$ on the left is 1 and the only possible $x^1$ term on the right is if $a = 1$ . Using these values of $a$ and $b$ we verify that the given differential equation is satisfied since we have

$\begin{align*} y'' &= \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!} x^{3n-2} \\[9pt] x^a y + b = xy &= x + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n+1} \\[9pt] &= x + \sum_{n=2}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3(n-1)-2)}{(3(n-1))!} x^{3(n-1)+1} &(\text{Reindexing})\\[9pt] &= \frac{1}{(3\cdot 1 - 2)!} x^{3 \cdot 1 - 2} + \sum_{n=2}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-5)}{(3n-3)!} \cdot \frac{(3n-2)}{(3n-2)} \cdot x^{3n-2} \\[9pt] &= \frac{1}{(3 \cdot 1 - 2)!} x^{3 \cdot 1 - 2} + \sum_{n=2}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!} x^{3n-2} \\[9pt] &= \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!} x^{3n-2}. \end{align*}$

Hence, we indeed have $y'' = x^a y + b$ .

Determine the interval of convergence of ∑xⁿ / (n!)² and show that it satisfies a given differential equation

by RoRi

April 19, 2016

Consider the function f(x) defined by the power series,

$f(x) = \sum_{n=0}^{\infty} \frac{x^n}{(n!)^2}.$

Determine the interval of convergence of f(x) and show that f(x) satisfies the differential equation

$xy'' + y' - y = 0.$

First, to determine the radius of convergence we use the ratio test,

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{n+1}}{(n+1)!^2} \right) \left( \frac{(n!)^2}{x^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x}{(n+1)^2} \\[9pt] &= 0. \end{align*}$

Therefore, the series converges for all $x$ (or the radius of convergence is $r = +\infty$ ). Next, to show that $f(x)$ satisfies the given differential equation we take the first two derivatives,

$\begin{align*} && y &= \sum_{n=0}^{\infty} \frac{x^n}{(n!)^2} \\[9pt] \implies && y' &= \sum_{n=1}^{\infty} \frac{nx^{n-1}}{(n!)^2} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n ((n-1)!)^2} \\[9pt] \implies && y'' &= \sum_{n=2}^{\infty} \frac{(n-1)x^{n-2}}{n ((n-1)!)^2} = \sum_{n=2}^{\infty} \frac{x^{n-2}}{n(n-1)((n-2)!)^2}. \end{align*}$

Plugging this into the given differential equation we have

$\begin{align*} xy'' + y' - y &= x \left( \sum_{n=2}^{\infty} \frac{x^{n-2}}{n(n-1)((n-2)!)^2} \right) + \left( \sum_{n=1}^{\infty} \frac{x^{n-1}}{n ((n-1)!)^2} \right) - \left( \sum_{n=0}^{\infty} \frac{x^n}{(n!)^2} \right) \\[9pt] &= \sum_{n=2}^{\infty} \frac{x^{n-1}}{n(n-1)((n-2)!)^2} + \sum_{n=1}^{\infty} \frac{x^{n-1}}{n ((n-1)!)^2} - \sum_{n=0}^{\infty} \frac{x^n}{(n!)^2} \\[9pt] &= \sum_{n=1}^{\infty} \frac{x^n}{(n+1)n((n-1)!)^2} + \sum_{n=0}^{\infty} \frac{x^n}{(n+1)(n!)^2} -\sum_{n=0}^{\infty} \frac{x^n}{(n!)^2} \\[9pt] &= \left( \sum_{n=1}^{\infty} \frac{x^n}{(n+1)n((n-1)!)^2} + \frac{x^n}{(n+1)(n!)^2} - \frac{x^n}{(n!)^2} \right) + 1 - 1 \\[9pt] &= \sum_{n=1}^{\infty} \left( \frac{1}{(n+1)n((n-1)!)^2} + \frac{1}{(n+1)(n!)^2} - \frac{1}{(n!)^2} \right)x^n \\[9pt] &= \sum_{n=1}^{\infty} \frac{n + 1 - (n+1)}{(n+1)(n!)^2} x^n \\[9pt] &= \sum_{n=1}^{\infty} \frac{0}{(n+1)(n!)^2} x^n \\[9pt] &= 0. \end{align*}$

Hence, $y$ indeed satisfies the given differential equation.

Determine the interval of convergence of ∑ x⁴ⁿ / (4n)! and show that it satisfies a given differential equation

by RoRi

April 19, 2016

Consider the function $f(x)$ defined by the power series,

$f(x) = \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}.$

Determine the interval of convergence of $f(x)$ and show that $f(x)$ satisfies the differential equation

$f^{(4)}(x) = f(x).$

First, to determine the radius of convergence we use the ratio test

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{4n+4}}{(4n+4)!} \right) \left( \frac{(4n)!}{x^{4n}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x^4}{(4n+1)(4n+2)(4n+3)(4n+4)} \\[9pt] &= 0. \end{align*}$

Therefore, $f(x)$ converges for all $x$ (equivalently, $r = +\infty$ ). Next, to show that $f(x)$ satisfies the differential equation $y^{(4)} = y$ we take the first four derivatives,

$\begin{align*} && y&= \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!} \\[9pt] \implies && y' &= \sum_{n=1}^{\infty} \frac{4n x^{4n-1}}{(4n)!} = \sum_{n=1}^{\infty} \frac{x^{4n-1}}{(4n-1)!} \\[9pt] \implies && y'' &= \sum_{n=1}^{\infty} \frac{(4n-1)x^{4n-2}}{(4n-1)!} = \sum_{n=2}^{\infty} \frac{x^{4n-2}}{(4n-2)!} \\[9pt] \implies && y''' &= \sum_{n=1}^{\infty} \frac{(4n-2)x^{4n-3}}{(4n-2)!} = \sum_{n=3}^{\infty} \frac{x^{4n-3}}{(4n-3)!} \\[9pt] \implies && y^{(4)} &= \sum_{n=1}^{\infty} \frac{(4n-3)x^{4n-4}}{(4n-3)!} = \sum_{n=4}^{\infty} \frac{x^{4n-4}}{(4n-4)!}. \end{align*}$

But, reindexing this expression for the fourth derivative we have

$y^{(4)} = \sum_{n=1}^{\infty} \frac{x^{4n-4}}{(4n-4)!} = \sum_{n=0}^{\infty} \frac{x^{4(n+1)-4}}{(4(n+1)-4)!} = \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!} = y.$

Thus, $f(x)$ satisfies the given differential equation.

Use the method of undetermined coefficients to solve (1-x²)y′′ – 2xy′ + 6y = 0

by RoRi

April 19, 2016

Consider the differential equation

$(1-x^2)y'' - 2xy' + 6y = 0.$

The solution to this differential equation has a power-series expansion

$f(x) = \sum_{n=0}^{\infty} a_n x^n \qquad \text{with} \qquad f(0) = 1, \quad f'(0) = 0.$

Using the method of undetermined coefficients obtain a recursion formula relating the terms $a_{n+2}$ to the terms $a_n$ . Give an explicit formula for $a_n$ for each $n$ and find the sum of the series.

First, we differentiate twice,

$\begin{align*} f(x) &= \sum_{n=0}^{\infty} a_n x^n \\ f'(x) &= \sum_{n=1}^{\infty} na_n x^{n-1} \\ f''(x) &= \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}. \end{align*}$

From the initial conditions $f(0) = 1$ and $f'(0) = 0$ we have

$\begin{align*} f(0) &= \sum_{n=0}^{\infty} a_n x^n = a_0 = 1 \\ f'(0) &= \sum_{n=1}^{\infty} na_n x^{n-1} = a_1 = 0. \end{align*}$

Now, we plug the expressions for $y, \ y'$ , and $y''$ back into the given differential equation,

$\begin{align*} && (1-x^2)y'' - 2xy' + 6y &= 0 \\[9pt] \implies && (1-x)^2 \left(\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} \right) - 2x \sum_{n=1}^{\infty} na_n x^{n-1} + 6 \sum_{n=0}^{\infty} a_n x^n &= 0 \\[9pt] \implies && \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - \sum_{n=2}^{\infty} n(n-1)a_n x^n - 2 \sum_{n=1}^{\infty} na_n x^n + 6 \sum_{n=0}^{\infty} a_n x^n &= 0 \\[9pt] \implies && \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=2}^{\infty} n(n-1)a_n x^n - 2 \sum_{n=1}^{\infty} na_n x^n + 6 \sum_{n=0}^{\infty} a_n x^n &= 0 \\[9pt] \implies && \left(\sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} - n(n-1)a_n - 2na_n + 6a_n \right)x^n\right) + 2a_2 + 6a_3 x -2a_1 x + 6a_0 + 6a_1 x &= 0. \end{align*}$

Then, we use the fact from above that $a_0 = 1$ and $a_1 = 1$ to get

$\begin{align*} && \left( \sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} - n(n-1)a_n - 2na_n + 6a_n \right) \right) + 2a_2 + 6a_3 x + 6 &= 0 \\[9pt] \implies && \left( \sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} - a_n (n^2 + n - 6) \right) \right) + 6a_3 x + 2a_2 + 6 &= 0 \\ \implies && \left( \sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} - (n+3)(n-2)a_n \right)\right) + 6a_3 x + 2a_2 + 6 &= 0. \end{align*}$

Since this sum is equal to 0, we know that every coefficient of every power of $x$ must be equal to 0. First, we solve for $a_2$ and $a_3$ ,

$\begin{align*} 2a_2 + 6 &= 0 & \implies && a_2 &= -3 \\ 6a_3 &= 0 & \implies && a_3 &= 0. \end{align*}$

Then, we establish the recursive relationship between $a_n$ and $a_{n+2}$ ,

$\begin{align*} && (n+2)(n+1)a_{n+2} - (n+3)(n-2)a_n &= 0 \\ \implies && (n+2)(n+1) a_{n+2} &= (n+3)(n-2)a_n \\ \implies && a_{n+2} &= \frac{(n+3)(n-2)}{(n+2)(n+1)} a_n \end{align*}$

for all $n \geq 0$ . Then since $a_3 = 0$ we have $a_{2n+1} = 0$ for all $n$ (since for every odd integer $n$ the formula for $a_{n+2}$ has is multiplied by $a_n$ , but each of these will be 0). For the even terms we have for $n=2$ ,

$a_4 = \frac{(n+3)(n-2)}{(n+2)(n+1)} a_2 = \frac{(5)(0)}{(4)(3)} (-3) = 0.$

This means all of the remaining even terms will be 0 as well. So we have

$a_0 = 1, \quad a_1 = 0, \quad a_2 = -3, \quad a_3 = a_4 = \cdots = 0.$

Hence,

$f(x) = 1 - 3x^2.$

Prove that a given differential equation has a given particular solution

by RoRi

February 27, 2016

Let $L$ be defined by

$L(y) = y'' + ay' + by,$

for $a, b \in \mathbb{R}$ constants.

Assume $c \in \mathbb{R}$ and $b \neq \omega^2$ . Prove that the differential equation

$L(y) = c \sin (\omega x)$

has a particular solution of the form

$y = A \sin (\omega x + \alpha)$

and find expressions for $A$ and $\alpha$ in terms of $a,b,c,$ and $\omega$ .

Proof. Incomplete.

Prove a given equation is a solution of a given differential equation

by RoRi

February 27, 2016

Let $L$ be defined by

$L(y) = y'' + ay' + by,$

for $a, b \in \mathbb{R}$ constants.

Let $c \in \mathbb{R}$ and $b \neq \omega^2$ . Using the previous exercise prove that the differential equation

$L(y) = c \cos (\omega x)$

has a particular solution given by

$y = A \cos (\omega x - \alpha),$

where

$A = \frac{c}{\sqrt{(b-\omega^2)^2 + a^2 \omega^2}} \quad \text{and} \quad \tan \alpha = \frac{a \omega}{b - \omega^2}.$

Proof. Incomplete.

Prove existence of a complex valued solution to a given differential equation

by RoRi

February 27, 2016

Let $L$ be defined by

$L(y) = y'' + ay' + by,$

for $a, b \in \mathbb{R}$ constants.

Let $\omega \in \mathbb{R}$ and $A \in \mathbb{C}$ . If either $b \neq \omega^2$ or $a \omega \neq 0$ prove that the differential equation

$L(y) = Ae^{i \omega x}$

has a solution

$y = Be^{i \omega x}.$

Express the value of $B$ in terms of $a$ , $b$ , $A$ , and $\omega$ .

Proof. Let $y = Be^{i \omega x}$ with either $b \neq \omega^2$ or $a \omega \neq 0$ . Then we have

$\begin{align*} L(y) &= -B \omega^2 e^{i \omega x} + a B i \omega e^{i \omega x} + bB e^{i \omega x} \\ &= e^{i \omega x} (bB - B\omega^2 + \omega a B i ). \end{align*}$

So,

$B = \frac{A}{b - \omega^2 + i a \omega} \qquad \text{if } b - \omega^2 + i a \omega \neq 0.$

But since $b - \omega^2 \neq 0$ or $a \omega \neq 0$ we have $b - \omega^2 + i a \omega \neq 0$ . Hence,

$L(y) = Ae^{i \omega x}.$

Deduce properties of the solutions of a given differential equation

by RoRi

February 24, 2016

Let $f(x)$ be a solution for all $x \in \mathbb{R}$ of the second order differential equation

$xf''(x) + 3x(f'(x))^2 = 1 - e^{-x}.$

Without attempting to solve the equation answer the following questions.

If $c \neq 0$ and $f$ has an extremum at $c$ , prove that this extremum must be a minimum.
If $f$ has an extremum at 0 decide whether this extremum is a maximum or minimum and prove your assertion.
If the solution $f(x)$ satisfies the conditions
$f(0) = f'(0) = 0$

find the minimum value for the constant $A$ such that

$f(x) \leq Ax^2 \qquad \text{for all } x \geq 0.$

Incomplete.