Detailed Explanations for Calculus 2 Equations

Undefined Terms Within Series Solution For Heat Equation Solution
$u_{t} = 4 u_{\times}, t > 0, 0 < x < π,$
$u_{x} (0, t) = u_{x} (π, t) = 0, t > 0$
$u (x, 0) = 2 - \cos (2 x) + 5 \cos (3 x), 0 \leq x \leq π$

Willie Kelley 2022-04-10

Trying to understand eigenvalues with respect to differential equations.
I am trying to understand how to find eigenvalues from a matrix consisting of exponential terms, considering a differential equation. The examples I've seen online are ODEs. Without using a vector with exponential terms, here is what I have learned.
$\frac{d}{d t} \vec{x} (t) = λ \vec{x} (t)$
$\frac{d}{d t} [\begin{matrix} x_{1} (t) \\ x_{2} (t) \\ x_{3} (t) \end{matrix}] = [\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix}] [\begin{matrix} x_{1} (t = 0) \\ x_{2} (t = 0) \\ x_{3} (t = 0) \end{matrix}]$
Here is what I am trying to understand
Section of a paper with imag eigenvalue
In this paper, the following assumption is made.
$\vec{J} (t) = | \vec{J} | e^{- i ω t} = [\begin{matrix} | J_{x} | e^{- i ω t} \\ | J_{y} | e^{- i ω t} \\ | J_{z} | e^{- i ω t} \end{matrix}]$
They are using partial derivatives (I believe this can be viewed as an ODE then?). Differentiating with respect to time I believe yields the following. Please correct me if I am wrong.
$\frac{\partial}{\partial t} \vec{J} (t) = \frac{\partial}{\partial t} [\begin{matrix} | J_{x} | e^{- i ω t} \\ | J_{y} | e^{- i ω t} \\ | J_{z} | e^{- i ω t} \end{matrix}] = [\begin{matrix} - i ω & 0 & 0 \\ 0 & - i ω & 0 \\ 0 & 0 & - i ω \end{matrix}] [\begin{matrix} | J_{x} | e^{- i ω t (t = 0)} \\ | J_{y} | e^{- i ω t (t = 0)} \\ | J_{z} | e^{- i ω t (t = 0)} \end{matrix}]$
Or
$\frac{\partial}{\partial t} [\begin{matrix} | J_{x} | e^{- i ω t} \\ | J_{y} | e^{- i ω t} \\ | J_{z} | e^{- i ω t} \end{matrix}] = [\begin{matrix} - i ω & 0 & 0 \\ 0 & - i ω & 0 \\ 0 & 0 & - i ω \end{matrix}] [\begin{matrix} | J_{x} | \\ | J_{y} | \\ | J_{z} | \end{matrix}]$
Are my assumptions correct? If so, is there a deeper analysis to why this is the case with exponential terms?

Dakota Livingston 2022-04-10

Third Order Differential Equation using the reduction of order method.
Solve the following differential equation $y^{‴} - 2 y^{″} + y^{'} - 2 y = 0$ (1) using the reduction of order method and also by replacing y with $z = z (x) = y^{'} - 2 y$ (2). So $z^{'} = y^{″} - 2 y^{'} (3), z^{″} = y^{‴} - 2 y^{″} (4)$ therefore (1) becomes $z^{″} + z = 0$ and we can easily find the general solution. So my question is... what about the function $z^{'} = y^{″} - 2 y^{'}$ .
We can not plug that in (1) because the function $z^{″} = y^{‴} - 2 y^{″}$ is the first two terms of the original differential equation and the function $z = y^{'} - 2 y$ is that last two terms.

Edith Mayer 2022-04-10

Integral $\int \frac{d x}{\sin x + \sec x}$

Magdalena Warren 2022-04-09

How do you solve the differential equation $\frac{d y}{d x} = 6 y^{2} x$ , where $y (1) = \frac{1}{25}$ ?

Blaine Jimenez 2022-04-09

How do you find the solution to the differential equation $\frac{d y}{d x} = \frac{\cos (x)}{y^{2}}$ where $y (\frac{π}{2}) = 0$ ?

Karla Thompson 2022-04-09

Use the superposition approach to solve the non-homogeneous differential equation,
$y^{″} + 6 y^{'} + 8 y = 4 x - e^{- 2 x}$

Cazzaoro0w9 2022-04-09

Use the annihilator method and find the general solution of $y^{″} - 3 y^{'} + 2 y = 4 \sin^{3} (3 x)$

Kasey Castillo 2022-04-09

Unable to solve the following question
I am trying to solve the following question but I can not figure out a method to solve it. The question is:
$x + \sqrt{(y^{2}) - (x y) \frac{d y}{d x}} = y, y (\frac{1}{2}) = 1$

Magdalena Warren 2022-04-09

Two questions about trig functions of matrices
So the first part of this question asked me to define cosA and sinA where A is a nxn real matrix. I used the taylor series expansions of sin and cos to get
$\sin A = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} A^{2 n + 1}$
$\cos A = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} A^{2 n}$
For part a. I need to show that
$\frac{d}{d t} \sin (A t) = A \cos (A t)$
$\frac{d}{d t} \cos (A t) = - A \sin (A t)$
I was able to do this for sin with the following steps:
$\frac{d}{d t} \sin (A t) = \frac{d}{d t} (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} {(A t)}^{2 n + 1}) = \frac{d}{d t} (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} A^{2 n + 1} t^{2 n + 1}) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} A^{2 n + 1} (2 n + 1) t^{2 n} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} A^{2 n + 1} t^{2 n} = A \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} {(A t)}^{2 n} = A \cos (A t)$
However, for cosine I get stuck. I start off the same way:
Calculus 2Answered question
Blaine Jimenez 2022-04-09
Trapping Region for ODE System.
$x^{'} = 2 x + y - 2 x^{3} - 3 x y^{2},$
$y^{'} = - 2 x + 4 y - 4 y^{3} - 2 x^{2} y,$
Calculus 2Answered question
Angelina Kaufman 2022-04-08
Finding the extrema of a functional (calculus of variations)
$I (y) = \int_{0}^{1} y^{' 2} - y^{2} + 2 x y d x$
Explanation:
$\begin{aligned} \frac{\partial F}{\partial y^{'}} & = 2 y^{'} ⟹ \frac{d}{d x} (\frac{\partial F}{\partial y^{'}}) = 2 y^{''} \\ \frac{\partial F}{\partial y} & = - 2 y + 2 x \end{aligned}$
Which gives the Euler equation $2 y^{″} + 2 y - 2 x = 0$ (I think, unless I messed up my math).
Ok now I need to try to find solutions to this differential equation, however, I only know how to solve linear second order DE's. The x term is throwing me off and I am not sure how to solve this.
I know $y (x) = x$ is a solution by inspection, but inspection is a poor man's approach to solving DE's.
Calculus 2Answered question
Haylee Bowen 2022-04-08
Finding the inverse Laplace transform after solving a differential equation
I wanted to solve $y^{″} + 9 y = \sin (3 x)$ , where $y (0) = 0$ and $y^{'} (0) = 0$ by doing a Laplace Transform.
Taking $L [y^{″} + 9 y] = L [\sin (3 x)]$ gives $s^{2} F (s) - s y (0) - y^{'} (0) + 9 F (s) = \frac{3}{s^{2} + 9}$
The initial values are both zero, and F(s) can be factored out of the left-hand side, leaving :
$F (s) = \frac{3}{{(s^{2} + 9)}^{2}}$ and I couldn't figure out how to inverse transform this, but I did see that it was similar to $L^{- 1} [\frac{2 a^{3}}{{(s^{2} + a^{2})}^{2}}] = \sin (a t) - a t \cos (a t)$ which might be helpful
So in the end, how do I do $L^{- 1} [\frac{3}{{(s^{2} + 9)}^{2}}]$ ?
Calculus 2Answered question
Kendal Day 2022-04-08
Finding solution of ODE using Laplace transform
with initial condition $y (0) = 0$ .
$y^{'} + 2 y = f (x)$ where $f (x) = 0 if x > 1 and f (x) = 1 if 0 \leq x \leq 1$ .
I applied Laplace transform on both sides which gets
$s Y (s) + 2 Y (s) = \frac{1 - e^{- s}}{s} \Rightarrow Y (s) = \frac{1 - e^{- s}}{s (s + 2)} = \frac{1}{s (s + 2)} - \frac{e^{- s}}{s (s + 2)}$
Now I know inverse Laplace transform of $\frac{1}{s (s + 2)}$ but how to find inverse Laplace transform of second term? Or can we solve above ODE using different method?
Calculus 2Answered question
Papierskiix5n 2022-04-08
I want to find the Taylor series for the function:
$f (x) = \ln (\tan (x)) - \ln (x)$
Calculus 2Answered question
Perla Benitez 2022-04-08
How to simplify $f (x) = \sum_{i = 0}^{\infty} \frac{x^{i mod (k - 1)}}{i!}$
Calculus 2Answered question
fanairana7lu1 2022-04-08
How to numerically solve this nonlinear ODE?
$u^{″} + \frac{2}{r} u^{'} - \frac{K}{{(1 + r^{2})}^{2}} u = u^{5}$
with initial conditions $u (0) = d > 0$ and $u^{'} (0) = 0$ .
Calculus 2Answered question
Destinee Bryan 2022-04-08
Using a change in variables, Show that the boundary value problem:
$- \frac{d}{d x} (p (x) y^{'}) + q (x) y = f (x),$
$a \leq x \leq b,$
$y (a) = α,$
$y (b) = β$
Can be transformed to the form:
$- \frac{d}{d w} (p (w) z^{'}) + q (w) z = F (w),$
$0 \leq w \leq 1,$
$z (0) = 0,$
$z (1) = 0$
Calculus 2Answered question
gabolzm6d 2022-04-08
Understanding how to apply dominant balance method
I have the differential equation
$ϵ y^{″} - x^{2} y^{'} - y = 0$
Calculus 2Answered question
Shaylee House 2022-04-08
Types of singularities in ODE
$y \cdot \frac{d^{2} y}{{d x}^{2}} - {(\frac{d y}{d x})}^{2} + 1 = 0$
with $y : R \to R$
As seen, it does not satisfy Picard–Lindelöf theorem because of singularity at $y = 0$ . Nevertheless it possesses smooth solutions, for example $y (x) = \frac{1}{a} \cos a x$ for $a \neq 0$ . On the other hand, all the solutions of
$y \cdot \frac{d^{2} y}{{d x}^{2}} + {(\frac{d y}{d x})}^{2} + 1 = 0$
explode when y approaches 0.
The first singularity leads to violation of uniqueness, while the second breaks the solutions.
Are there special terms for these types of singularities? A study? The term "regular singular point" seems to have a different meaning.

When you are dealing with any Calculus 2 homework, it is vital to have a look at the various questions and answers that will help you see whether you are correct in your approach to finding solutions. Even if you are dealing with analytical aspects of Calculus 2, it will be helpful as you are looking at provided equations and learn how the answers relate to original questions and problems specified.

Do not be afraid to take a look at the basic integration and related application if Calculus 2 does not sound clear or start with the Calculus 1 first.

Calculus 2

Calculus 2 Equations: Expert Solutions and More