# Tag Info

0

Hint: Assume that $$y = \sum_{n=0}^\infty a_nx^n.$$ Calculate the first and 2nd derivatives, express $e^{-x}$ also as a power series. Plug them all into the equation. Make use of this identity: if $\sum_{n=0}^\infty b_nx^n = 0$, then $b_n = 0$ for all $n$.

0

Taking another example $$\left\{\begin{matrix} L_1y=ay'+by=0\\ L_2y=cy''+dy'+ey=0 \end{matrix}\right.$$ now this becomre $$\left\{\begin{matrix} L_1y=ay'+by=0\\ L_2'y=ay''+by' + cy''+dy'+ey=(a+ c)y''+(b +d)y'+ey=0 \end{matrix}\right.$$ Note that in this case $L'$ is doesn't really "better" then $L$. When you talking about linear system of ...

0

Yes, you can look at the (partial) linearization in the $s_i$ variables. In your case this is relatively easy because $s_1 = s_2 = s_3 = 0$ is always a fixed point (or actually, invariant set) of the system. Secondly, the linearization in the $s$ variables at zero is precisely given by the diagonal matrix $$A = \textrm{diag}(\frac{\partial f_1}{\partial ... 0 y=c_1e^{3x}\pmatrix{1\\1}+c_2e^{4x}\pmatrix{1\\2}. y'=3c_1e^{3x}\pmatrix{1\\1}+4c_2e^{4x}\pmatrix{1\\2}. y'(0)=3c_1\pmatrix{1\\1}+4c_2\pmatrix{1\\2}. so y'_2(0)=3c_1+8c_2=3 and we got c_1=1-\frac{8}3c_2 Now, e^{-4x}y(x)= c_1e^{-x}\pmatrix{1\\1}+c_2\pmatrix{1\\2}= (1-\frac{8}3c_2)e^{-x}\pmatrix{1\\1}+c_2\pmatrix{1\\2} and e^{-4t}y_1(t)= ... 2 If you read a real analysis textbook such as Calculus by Spivak, they manage to develop calculus rigorously while avoiding differentials like "dx" and "dy" entirely. This is the standard way to make calculus rigorous -- you just avoid using differentials. And indeed, in undergrad differential equations classes, arguments that involve manipulating dx ... 1 To find the particular solution, you simply take your general solution and plug in the values that you are given for the particular solution. Your general solution is$$y=Ae^x+Be^{2x}+2\sin x+6\cos x.$$You have given that the particular solution has the properties y(0)=0 and \frac{dy}{dx}(0)=0. The first condition means that when x=0, then y=0, ... -1 Hint: Depending on undetermined Coefficients method and on the R.H.S, the particular solution should be$$y_p=C_1\cos x+C_2\sin x$$The particular solution of non-homogeneous differential equation not depend on the initial conditions as shown$$y'_p=-C_1\sin x+C_2\cos xy''_p=-C_1\cos x-C_2\sin x$$substitute them in D.E$$-C_1\cos x-C_2\sin ...

0

I think a differential equation is homogeneous if every term contains y or derivatives of y in the equation

0

To combine the (very useful!) comments of @Did and @Evgeny in an answer: The definition of reversibility I'm using is that of time reversal symmetry, i.e. the motion described by the system backward in time is equivalent to that forward in time (I'm following James D. Meiss, Differential Dynamical Systems, SIAM MM14, 2007, p. 212 here). That means that ...

3

Notice, the given D.E. can be easily solved by exact differential form as follows $$(y+1)dx+(x+y)dy=0$$ Now, comparing the above equation with $Mdx+Ndy=0$, we get $$M=y+1\implies \frac{\partial M}{\partial y}=1$$ $$N=x+y\implies \frac{\partial N}{\partial x}=1$$ since, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, hence the given equation ...

-3

the answer should be $$y(x)=x\pm \sqrt{x^2+2x+C}$$

6

See that we can write as $$(x+y)dy+(y+1)dx=0$$ $$xdy+ydx+ydy+dx=0$$ $$d(xy)+ydy+dx=0$$ now just integrate.

0

Sort of guide Transform the equation such that it'll have homogeneous boundary conditions. I suggest to use $W(x, y) = 100 + 50x$ : this is the simplest function that has $W(0, y) = 100$, $W(2, y) = 200$ and by the way $\frac{\partial W}{\partial y} \equiv 0$. What will happen to solutions of original equation if we subtract $W(x, y)$ ? Let's see: $$\Delta ... 1 First, observe that $$\frac{\partial^2}{\partial x \partial y} \frac{1}{f} = \frac{1}{f^2}\left(2 \frac{f_x f_y}{f} - f_{x y}\right),$$ so if \alpha = 2, this takes us a step closer to the solution. The above inspires us (at least, it inspired me) to try \frac{\partial^2}{\partial x \partial y} \frac{1}{f^\beta} ... 0 Hints 1) Start from here$$Y^{''} - \lambda Y = 0\\ Y^{'}(0)=Y^{'}(4)=0$$find the eigenvalues \lambda_i and the eigenfunctions Y_i(y). 2) Find the X_i(x)$$X^{''} + \lambda X = 0$$3) Form the infinite sum$$T(x,y) = \sum\limits_{i = 1}^\infty {{X_i}(x){Y_i}(y)} $$4) Apply boundary conditions$$\eqalign{ & \sum\limits_{i = ...

4

(This integral also comes up in the brachistochrone problem, by the way.) Rearranging it into an integrable form gives $$\frac{\sqrt{R}}{\sqrt{-2GM/C-R}} \dot{R} = \sqrt{-C},$$ so set $K=-2GM/C$. Integrating both sides, $$\int_0^R \frac{\sqrt{r}}{\sqrt{K-r}} \, dr = \sqrt{-C}t.$$ Doing the substitution $r=K(1-u^2)$, so $dr=-2Ku\, du$ the integral ...

0

If you want to know what a mathematical thing is, you need two things: How it works - what the rules of using it are. A way of expressing it in terms of mathematical things you know and trust. For differentials, we know what the rules of using them are. And I'm told some clever folks have worked out how to model them (notably Abraham Robinson with ...

1

Since $$\frac{dy}{dx}-\frac{dx}{dy}=\frac{dy}{dx}-\frac{1}{\frac{dy}{dx}}=\frac{x}{y}-\frac{y}{x}$$ We have $$\frac{dy}{dx}=\frac{x}{y}$$ or $$\frac{dy}{dx}=-\frac{y}{x}$$ If $\frac{dy}{dx}=\frac{x}{y}$, we have $y dy = x dx$, so integration gives $y^2=x^2+C$, or $y=\pm \sqrt{x^2+C}$. If $\frac{dy}{dx}=-\frac{y}{x}$, we have $-\frac{dy}{y}=\frac{dx}{x}$, ...

1

Hint Write, from definition, $$y=\sum_{n = 0}^\infty a_nx^n=\sum_{n = 0}^\infty \frac{y^{(n)}}{n!}x^n$$ Now, the differential equation $$y' = x − y^2$$ and we continue differentiating $$y''=1-2y\,y'$$ $$y'''=-2y'^2-2y\,y''$$ $$y^{(4)}=-6y'y''-2y\, y'''$$ and so forth. I am sure that you can take from here.

3

In both cases first verify the claim for $\phi(D)=D,$ then by induction for $\phi(D)=D^n$ (this is the most interesting step), then by linearity for a general polynomial $\phi.$ Or if you are as lazy as I am, your first step could even be the constant $\phi(D)=1.$

0

In order to show that $h(x)=O(\| x \|^3)$ use the Taylor expansion around the origin; $$h(x)=-\frac{a}{2} x_1^3(ax_2-\frac{a}{6}x_2x_1^3-\frac{x_1^2}{2}+h.o.t)$$ Hence, in a bounded neighborhood of the origin one gets: $$|h(x)|\leq k|x_1|^3\leq k \|x\|^3$$ for some $k\in \mathbb{R}^{+}$. To show that $\dot{V}$ is negative definite, first we prove that the ...

0

A set of functions is linearly dependent on an interval $I$ if the Wronskian is $\textit {identically}$ zero there. In your case, if you pass to the characteristic equation $r^4-1=(r^2+1)(r-1)(r+1)$, you can read off the fundamental set: $\left \{ \cos t,\sin t, e^{t},e^{-t} \right \}$. Now a calculation shows that the Wronskian is not identically zero on ...

3

If a set of functions are linearly dependent, the Wronskian will be zero everywhere. That is, for all $t$, we will have $e^t\cos(t)-e^t\sin(t)=0.$ That this is not the case at $t=0$ tells us the functions are independent. Moreover, the Wronskian being zero everywhere is not a sufficient condition for functions to be linearly dependent - Wikipedia gives $x^2$ ...

0

Never mind! I found the answer, I think. If we call $r/2=a$ in the above expression, and substitute $m=n-a+1$, then we get something like: $$\lim_{a \to \infty} \left[ \sum_{n=0}^{a} \frac{\hbar^n (-1)^n}{n!} \sum_{k=0}^n \binom{n}{k} \sum_{i=k}^{2a-n} \frac{x^{i-k}}{(i-k)!} \frac{d^{n-k}}{dx^{n-k}} + \sum_{m=0}^{a+1} \frac{\hbar^{m+a-1} ... 0 Apparently, the operation that is performed is this: the coefficients of the polynomials a_i(x) belong to the ring of algebraic integers A of an algebraic number field K. For any prime ideal \mathfrak p of A, we can consider the quotient ring A/\mathfrak p, which is an integral domain and has a field of fractions, which is called the residue ... 0 \begin{bmatrix} a_{1}\\ b_{1} \\ \end{bmatrix} =\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} \begin{bmatrix} a_0\\ b_0 \\ \end{bmatrix} + \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}= \begin{bmatrix} 5a_0 + b_0+1\\ -a_0+3b_0 -1 \end{bmatrix}  \begin{bmatrix} a_{2}\\ b_{2} \\ \end{bmatrix} =\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} ... 0 Assume$$y(x)=\sum_{n=0}^{\infty} a_n x^n,$$then$$y'(x)=\sum_{n=0}^{\infty} a_n n x^{n-1}$$and$$y''(x)=\sum_{n=0}^{\infty} a_n n (n-1)x^{n-2}$$substitute to$$xy′′+2y′+(λ^2) \, xy=0$$to get$$\sum_{n=0}^{\infty} a_n n (n-1)x^{n-1}+2\sum_{n=0}^{\infty} a_n n x^{n-1}+(λ^2) \, \sum_{n=0}^{\infty} a_n x^{n+1}=0\sum_{n=1}^{\infty} a_{n+1}( n+1) n ...

1

In order to find the solutions to the homogenous differential equation , solve the characteristic equation: $$\lambda^2+2 \lambda+5=0$$

0

Rate of change is of a function $y=f(x)$ is a ratio of change in $y$ vs change in $x$, i.e. $\frac{y_2-y_1}{x_2-x_1}\equiv\frac{\Delta y}{\Delta x}$ if you look at a momentary rate of change you get a derivative $$\frac{df}{d x}=\lim_{x_2\to x_1}\frac{y_2-y_1}{x_2-x_1}=\lim_{x_2\to x_1}\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ If $f(x)$ is a linear function, then ...

1

Let $x$ be an integral curve through $p$: $$x'(t) = V\bigl(x(t)\bigr),\quad x(0) = p.$$ Since $V$ is differentiable at $p$ and $V(p) = 0$, there exists an $M > 0$ such that $\|V(x)\| \leq M \|x - p\|$ for all $x$ in some neighborhood of $p$. By the flow equation and the triangle inequality, \begin{align*} \|x(t) - p\| ...

1

By theorems on existence and uniqueness for analytic solutions of differential equations, the solution should be analytic at $0$, so there will be such an expansion. Since $y(0) = 0$, the series starts $y(x) = a_1 x + \ldots$. That would make $y'(x) = a_1 + \ldots$ and $e^{y(x)} = 1 + y(x) + \ldots = 1 + \ldots$, so the order $0$ expansion of your d.e. ...

2

Here is another way of solving the equation: $$y'(x)+x+e^{y(x)}=0\Longleftrightarrow$$ $$\frac{\text{d}y(x)}{\text{d}x}+x+e^{y(x)}=0\Longleftrightarrow$$ Let $y(x)=\ln(v(x))$, which gives $\frac{\text{d}y(x)}{\text{d}x}=\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)}$: $$x+v(x)+\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)}=0\Longleftrightarrow$$ ...

3

Yes, you can do it. We know $y(0)=0$. From the equation we get $y'(0)=-e^{y(0)}=-1.$ Derivate the equation: $$y''+1+e^y\,y'=0\implies y'(0)=0.$$ Derivate once more to find $y'''(0)$ and still once more for $y''''(0)$. There is another way of solving the equation. The term $y'+x$ suggests the change $z=y+x^2/2$. This changes the equation into an equation ...

21

Let $y'=a(x)\,y+f(x)$ be the equation. We have to determine $a$ and $f$. The function $y_1-y_2$ is a solution of the homogeneous equation $y'=a\,y$, that is $$2\,x+\frac{1}{x^2}=a(x)\Bigl(x^2-\frac1x\Bigr).$$ From this you find $a$, and then $f$. 

1

The most general approach to these problems is to write your system as a four dimensional first order ODE system: \begin{align} x' &= \xi\\ \xi' &= - b x - a \xi +d y + c \eta\\ y ' &= \eta\\ \eta' &= -d x -c \xi -b y -a \eta \end{align} which can be written in matrix form $$\mathbf{x}' = A\, \mathbf{x}$$ with ...

0

You already have $$y(t) = \frac{1}{t^2-1+1/y_0}.$$ Check the denominator : $t^2-1+1/y_0$ have solution in $\mathbb{R}$ if and only if $1 - 1/y_0 \geqslant 0$, i.e. $y_0 \geqslant 1$ or $y_0 < 0$. If $y_0 \geqslant 1$ or $y_0 < 0$, let $t_0 = -\sqrt{1 - \frac{1}{y_0}}$. When $t$ moves from $-1$ to $t_0$, the value of $y$ tends to $+\infty$. So $y$ ...

2

About question 1. I have several comments here: It is strange for me to request necessary conditions. I would rather ask for sufficient conditions. In particular if you have in mind to apply Picard–Lindelöf theorem. For example the IVP problem $$y^\prime=1+y^{\frac{2}{3}}, y(0)=0$$ is such that $\frac{\partial f}{\partial y}$ doesn't exist at the origin. ...

2

Your equation blows up at $-1$ and $1$. Your interval of definition has to contain $0$ because that's where your initial condition is. So you will solve on $(-1,1)$. There, $\frac{t+1}{t-1}$ is always negative, and its absolute value is $\frac{t+1}{1-t}$.

0

As Dylan wrote, $y'$ is $\frac{dy}{dx}$. There is no differentiating with respect to $r$. What you're doing is guessing a solution of the form of an answer $y=e^{rx}$. Here, $r$ is some unknown that you wish to eventually find to complete your solution. It doesn't matter what you call it. I could call it $y=e^{\lambda x}$ where $\lambda$ is my parameter ...

0

You're solving for $y$ as a function of $x$ and $y' = \frac{dy}{dx}$ in the equation. $r$ is just a parameter.

2

Let us solve this problem in detail to show what kind of problems we encounter in here. We insert the ansatz $y(x) = \sum\limits_{n=0}^\infty p_n x^{n+\alpha}$ into the ODE and then equate to zero the coefficients at consecutive powers. We have: \begin{eqnarray} coeff @ x^{\alpha-2} :&& p_0 \alpha (\alpha-1) + a p_0 =0 \\ coeff @ ...

1

the general form of first order liner D.E $$y'+P(x)y=Q(x)$$ so that $P(x)=-\frac{4}{3}$ and $Q(x)=-4$ $$\rho=e^{\int P(x)dx}=e^{-\frac{4}{3}x}$$ $$\rho.y=\int \rho Q(x)dx=\int e^{-\frac{4}{3}x}(-4)dx=3e^{-\frac{4}{3}x}+C$$ hence $$y=3+Ce^{\frac{4}{3}x}$$

0

$$y=\sin 2x + 2\sin x \quad,\quad x \in [0,2\pi]$$ To find the extreme points we consider all critical points. These come from three sources. 1) From the first derivative test, requiring $y'=0$, we have three critical points as follows. $$y' = 2\cos 2x + 2\cos x = 0$$ $$\cos 2x + \cos x = 0$$ Using $\cos 2x = 2\cos^2 x - 1$, we have $$2\cos^2 x + \cos x ... 2 Based on your description of the problem, I believe that your ODE for the combustion byproducts volume C(t) is correct -- assuming that the gases don't cool off, which will have an effect on the density. So, we have $$C'(t) = r - \frac{r}{V} C(t).$$ This is not equivalent with C' + C = V, since C' + C = r + C ... 0 You could have negative values for K in your solutions. There are two branches to this DE. If you had a suitable initial condition to the DE then wolfram (I don't have maple) find it, e.g.: http://www.wolframalpha.com/input/?i=dy%2Fdx%3D%28y%5E2-1%29%2F%282y%29%2C+y%280%29%3D-1%2F2 As mentioned in the comments a complex value of C could give rise to a ... 0 You are right, except that you should be more careful with the constant solutions y=\pm 1. (In the final answer, they correspond to K=0, but you have defined K=\pm e^C, which can never be zero, so you have to "repair" the solution by including those cases "by hand".) 0 This is a linear ODE with constant coefficients of order 2. So start with m^2-1=0 as the method suggested and then make a linear combination of the two correspondending solutions. 1$$x''-x=0$$the characteristics equation is$$r^2-1=0r=\pm 1$$so$$x=c_1e^t+c_2e^{-t}$$0 Hint : \cosh(t) and \sinh(t) are two linear independent solutions. You can also take e^t and e^{-t} 0 I don't think your professor's procedure simplifies things. In the first place we have to establish a formula relating differentiation with respect to x (denoted by a {}'\>) with differentiation with respect to t (denoted by a \dot{}\>), valid for any function$$u:\quad {\mathbb R}_{>0}\to{\mathbb R},\qquad x\mapsto u(x)\ , resp., its ...

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