Detailed Explanations for Calculus 2 Equations

Computing the sum of the ratios:
$S_{n} = \sum_{k = 0}^{n} \frac{n - k + 1}{(n + 1) (n + 2) (n - k + 2)}$

Pizzadililehz 2022-03-27

How do I get an estimate for this nonlocal ODE?
Consider the following nonlocal ODE on $[1, \infty)$ :
$r^{2} f^{″} (r) + 2 r f^{'} (r) - l (l + 1) f (r) = - \frac{(f^{'} (1) + f (1))}{r^{2}}$
$f (1) = α$
$lim_{r \to \infty} f (r) = 0$
where l is a positive integer and $α$ is a real number.
Define the following norm $‖ \cdot ‖$
$‖ f ‖^{2} : = \int_{1}^{\infty} r^{2} f' (r)^{2} d r + l (l + 1) \int_{1}^{\infty} f (r)^{2} d r$
I want to prove the estimate:
$‖ f ‖ \leq C \sqrt{l (l + 1)} | α |$
for some constant C independent of $α$ , l and f. But I am stuck.
Here is what I tried. Multiply both sides by f and integrate by parts to get:
$\begin{aligned} ‖ f ‖^{2} = \int_{1}^{\infty} r^{2} f^{2} + \int_{1}^{\infty} l (l + 1) f^{2} & = f^{'} (1) f (1) + (f^{'} (1) + f (1)) \int_{1}^{\infty} \frac{f}{r^{2}} \\ \leq f^{'} (1) α + \frac{(f^{'} (1) + α)}{3} \sqrt{\int_{1}^{\infty} f^{2}} \\ \leq f^{'} (1) α + \frac{(f^{'} (1) + α)}{3} ‖ f ‖ \end{aligned}$
where I used Cauchy-Schwartz in the before last line. I am not sure how to continue and how to get rid of the f'(1) term.
Any help is appreciated.

amonitas3zeb 2022-03-27

How can I solve this differental equation?
$\begin{aligned} (\frac{1}{x} - y^{2} \frac{1}{(x - y)^{2}}) d x & + (x^{2} \frac{1}{(x - y)^{2}} - \frac{1}{y}) d y = 0 ? \end{aligned}$
I have this as a part of my homework, and there were like 20 other differential equations that I easily solved, but this one stood out.
This is not linear, not separable. I don't know how to approach this problem. If you could tell me to what class this kind of equation belongs, and some methods to solve them, I would be very very glad.

Sage Fry 2022-03-27

How am I supposed to find the solution to the non-homogeneous ode ?
$\vec{Y}' = (\begin{matrix} - 4 & 2 \\ 2 & - 1 \end{matrix}) \vec{Y} + (\begin{matrix} x^{- 1} \\ 2 x^{- 1} + 4 \end{matrix})$

metalskaashw 2022-03-27

Heat transfer between two fluids through a sandwiched solid (coupled problem):
Two fluids $(t_{h}, t_{c})$ flow opposite to each other on either side of a solid (T), while exchanging heat among themselves. In such a scenario, the conduction in the solid is governed by:
$x \in [0, 1], y \in [0, 1]$
1) $κ \frac{d^{2} T}{d x^{2}} + μ b_{h} (t_{h} - T) - ν b_{c} (T - t_{c}) = 0$
with boundary condition as $T^{'} (0) = T^{'} (1) = 0$
The fluids are governed by the following equations:
2) $\frac{d t_{h}}{d x} + b_{h} (t_{h} - T) = 0$
3) $\frac{d t_{c}}{d x} + b_{c} (T - t_{c}) = 0$
The hot fluid initiates at $x = 0$ and the cold fluid starts from $x = 1$ .
The boundary conditions are $t_{h} (x = 0) = 1$ and $t_{c} (x = 1) = 0$
Equation (1), (2) and (3) form a coupled system of ordinary differential equations.
It is pretty evident that using (2) and (3), Equation (1) can be re-written as:
4) $κ \frac{d^{2} T}{d x^{2}} - μ \frac{d t_{h}}{d x} + ν \frac{d t_{c}}{d x} = 0$
However, I have not been able to proceed further.
Some parameter values are
$b_{c} = 12.38, b_{h} = 25.32, μ = 1.143, ν = 1, κ = 2.16$

calcolare45pj 2022-03-27

Help calculating a series
$\sum_{n = 0}^{4} 2 n$

calcolare45pj 2022-03-27

Given that f(z) is a strictly increasing function (with real input and output and no root) for which there exists some positive (real) value of b such that it asymptotically outgrows $z^{b + 1}$ for positive z.
Is it true that any solution to
$\frac{d y}{d x} = f (y)$
will contain a vertical asymptote providing y (real) exceeds 0 at some point?
Conversely, is such a vertical asymptote impossible given that $z^{b + 1}$ outgrows f(z)?

brunocazelles4dvb 2022-03-26

Differential equation has exactly one solution
Let $n, d \in N$ and $τ > 0$ .
$f : [0, n τ] \times R^{d} \times R^{d} \to R^{d}, (t, x, y) \to f (t, x, y)$ continuous with the condition: There exists a $L > 0$ so that $| f (t, x_{1}, y) - f (t, x_{2}, y) | \leq L | x_{1} - x_{2} | \forall t \in [0, n τ], x_{1}, x_{2}, y \in R^{d}$ .
Furthermore, let $ϕ \in C^{1} ([- τ, 0], R^{d})$ .
${\begin{cases} y^{'} (t) = f (t, y (t), y (t - τ)), t \in [0, n τ], \\ y (t) = ϕ (t), t \in [- τ, 0] \end{cases}$
with the condition $ϕ^{'} (0) = f (0, ϕ (0), ϕ (- τ))$ . Show, that this problem has exactly one solution $y \in C^{1} ([- τ, n τ], R^{d})$ .

Hugh Soto 2022-03-26

Consider the following ODE
$x^{″} + x + x^{3} = 0$
Show that it only has periodic solutions.

Oxinailelpels3t14 2022-03-26

Consider an ODE system
$\dot{x} = f (x)$ ,
having a candidate Lypunov function, which satisfies $V (x) \geq 0, V (0) = 0$ , and $\dot{V} (x) \leq 0$ .
How to show that $Ω_{c} = {x \in ℝ^{n} : V (x)}$ is compact?

tibukooinm 2022-03-26

How can I handle $e x p (\ln | x |)$ to solve 1st order linear DE?
RHS and LHS are same.
$\exp (\ln (x)) = \exp (\ln (x)) (1)$
Taking log.
$\ln (\exp (\ln (x))) = \ln (\exp (\ln (x))) (2)$
$\ln (x) = \ln (\exp (\ln (x))) (3)$
$∴\approx x = \exp (\ln (x)) (4)$
Then what about ln|x| ?
This problem is a sub-problem of below ODE.
$\frac{y}{x} - \frac{2}{x} y = x^{6} (5)$
My thoughts are as below.
$P (x) = - \frac{2}{x} (6)$
$Q (x) = x^{6} (7)$
$\underset{1st order linear DE}{\underset{⏟}{\frac{y}{x} + P (x) y = Q (x)}} (8)$
$Integrating factor = \exp (\int P (x) dx) (9)$
$\exp (\int - \frac{2}{x} dx) (10)$
$= \exp (- 2 \int \frac{1}{x} dx) (11)$
$= \exp (- 2 \ln | x | + const) (12)$
$= \exp (- 2 \ln | x |) \exp (const) (13)$
$= {\exp (\ln | x |)}^{- 2} \exp (const) (14)$
$\exp (\ln | x |) = \exp (\ln | x |) (15)$
I got the following general solution as I forcefully assumed
$~ x = \exp (\ln | x |) ~$
$y = \frac{1}{5} x^{7} + {const}_{1} x^{2} (16)$

Erik Cantu 2022-03-26

Homogenous second order differential equation that seems not to be solvable
$(x^{2} + y^{2}) d x + (x^{2} - x y) d y = 0$
When I check for it to have a common partial derivative, I find that
$\frac{\partial M}{\partial y} = {(x^{2} + y^{2})}_{y} = 2 y$
and the second $\frac{\partial N}{\partial x} = {(x^{2} - x y)}_{x} = 2 x - y$
Thus, it does not seem to have common partial derivative. What can be done to solve it?
UPDATE:
by Moo's suggestion we get:
$(1 + u^{2}) d x + (1 - u) (x d u + u d x) = 0$
$(1 + u^{2}) d x + (x d u + u d x - u x d u - u^{2} d x) = 0$
$(1 + u) d x + x (1 - u) d u$
$\frac{1 + u}{1 - u} d x = - x d u$
this gives:
$\ln x + c = - \int \frac{1}{1 + u} + \frac{u}{1 + u} d u$
$\ln x + c = - \ln | 1 + u | + \int 1 d u - \int \frac{1}{1 + u} d u$
$\ln x + c = u - 2 \ln | u + 1 |$
Insert for $u = \frac{y}{x}$
$\ln x + c = - 2 \ln | \frac{y}{x} + 1 | + \frac{y}{x}$

amantantawq5l 2022-03-26

Given: $x y^{^{″}} + (2 + 2 x) y^{^{'}} + 2 y = 0$ has a solution $y_{1} (x) = x^{- 1}$ , find the general solution of the differential equation $x y^{^{″}} + (2 + 2 x) y^{^{'}} + 2 y = 8 e^{2 x}$

Jaylin Clements 2022-03-26

Given the following series: $\sum_{n = 1}^{\infty} {(\sqrt[3]{n + 1} - \sqrt[3]{n - 1})}^{α}$ where $α \in R$ . Does the series converge or diverge?

Pizzadililehz 2022-03-25

Differential equations
${(x + 1)}^{2} y^{″} + (x + 1) y^{'} + y = x^{2} + 2 \sin (\ln (x + 1)), y (0) = \frac{1}{5}, y^{'} (0) = 2$

Jasmine Todd 2022-03-25

Continuous function composed with itself is equal to propagation of a differential equation.
This question has been bugging me for a while. It was given as the last question of a first year undergrad analysis exam and so should be solvable with little machinery, yet it seems to point straight at ODEs which have yet to be covered. Here is the question:
Let $F : R \to R$ be a bounded $C^{1}$ function and ${(f_{t} : R \to R)}_{t \in R}$ a family of continuous functions with $f_{0} (x) = x$ such that
$lim_{h \to 0} \frac{f_{t + h} (x) - f_{t} (x)}{h} = F (f_{t} (x))$
Show that there exists a continuous function $g : R \to R$ such that $f_{1} = g \circ g$ .
I have attempted this with a classmate and we've come to various levels of understanding of the question, but the required conclusion keeps escaping us. Our main issues with the question are that we have very little understanding on continuous functions composed with themselves, and additionally we haven't managed to use the $C^{1}$ and boundedness condition on F.
Note that the question also states that we are allowed to assume that the family of functions ${(f_{t})}_{t \in R}$ is uniquely determined by the given conditions: this again points to the domain of differential equations, which ideally we should not need to refer to in order to solve this.

Dexter Odom 2022-03-25

Consider the problem
$Y^{'} = d \frac{1}{t + 1} + c \cdot \tan^{- 1} (Y (t)) - d \frac{1}{2}, Y (0) = 0$
with c a given constant. Since $Y^{'} (0) = d \frac{1}{2}$ , the solution Y (t) is initially increasing as t increases, regardless of the value of c. As best you can, show that there is a value of c, call it c∗, for which (1) if $c > c \cdot$ , the solution Y(t) increases indefinitely, and (2) if $c < c \cdot$ , then Y(t) increases initially, but then peaks and decreases. Using ode45, determine c∗ to within 0.00005, and then calculate the associated solution Y(t) for $0 \leq t \leq 50$ .

basura8w081 2022-03-25

Compute the linearization of an operator representing the Van der Pol oscillator
Let $G^{μ} : C^{\infty} (R) \to C^{\infty} (R)$ be an operator defined as
$G^{μ} [x] = x^{″} + μ (x^{2} - 1) x^{'} + x$
Where $C^{\infty} (R)$ is the space of infinitely-differentiable functions defined on $R$ . The case $G^{μ} = 0$ describes the Van der Pol oscillator.
Define the linearization of $G^{μ}$ at x to be the operator:
$L_{x}^{μ} [v] = lim_{h \to 0} \frac{G^{μ} (x + h v) - G^{μ} (x)}{h}$
Question: compute $L_{x}^{μ} [v]$ for any x, $v \in C^{\infty} (R)$ and $μ \in R$ .
The operator $L_{x}^{μ} [v]$ looks similar to the limit definition of the derivative, but I don't think I am simply meant to differentiate $G^{μ}$ . Finding $G^{μ} (x + h v) - G^{μ} (x)$ just gives a really long expression that doesn't look like it simplifies to anything useful.

dotzis16xd 2022-03-25

How do I find a second order differential equation given its solution?
Given that $y = A e^{x} + B e^{- 2 x} - x$
Find $a y^{″} (x) + b y^{'} (x) + c y = f (x)$
I have considered the auxiliary equation whose roots should be the exponents in the given equation: $k_{1} = 1 and k_{2} = - 2$ . I believe that $f (x) = p x + q$ given the additional -x term but I don't know how to find the values of a,b or c for the differential equation.

Lorelei Stanton 2022-03-25

How do I calculate the sum of this series?
$\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n 2^{n + 1}}$

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