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0

You have: $(A-\lambda)a = (k-1)v$. Now, $\psi(t) = a \exp(\lambda t)$, $\psi'(t) = a \lambda \exp(\lambda t)$, we have: $ \lambda a \ne Aa + v \ \forall a \in \mathbb{R}^n$, so for any vector $a\in R^n$: $(A-\lambda)a \ne -v$, so $k-1\ne -1$, i.e $k\ne 0$. So you found yourself suitable $a,b$, $b =kv , k \ne 0$, $a$ satisfies: $(A-\lambda)a = (k-1)v$.


0

Yes, if the ODE satisfies the Lipschitz condition so that there exists a unique exact solution, then the Euler polygon of stepsize $h$ has an error of size $\sim \frac{M}L·(e^{Lt}-1)h$ at time $t$. $M$ is a bound on $f$ and $L$ is the Lipschitz constant of $f$.


1

you can integrate both sides to get a first differential equation as follow: $$y'+2y=t+t^3/3-1/2e^{-2t}+C$$ and you can find the $C$ from the boundary condition


0

First calculate the solution of the homgeneous problem $y''+2y'=0$, the characteristic equation is $m^2+2m=0\Rightarrow m=0,-2$, so $y_c=A+Be^{-2t}$. We want to find $y=y_c+y_p$ that solves the full problem, so we just need to find $y_p$, if the right hand side contains a polynomial of some degree, then the ansatz must contain a polynomial of that degree as ...


0

The general solution $y$ is equal to $y_c+y_p$, where $y_c$ is the solution of the homogeneous equation, $y_p$ is a particular solution of the nonhomogeneous equation. For $y_c$, set up the characteristic equation $r^2+2r=0$. This will give you $y_c$. For $y_p$, separate the right hand side into two parts $1+t^2$ and $e^{-2t}$ and use superposition ...


1

I have written a long Section of a Chapter on this, in my ODEs book, but it is in Greek. If you understand I can send it to you. The spirit is the following: How to make separation of variables rigorous. However, I wish to make a comment. There are solutions which do not belong in your sets when uniqueness is violated. For example: Let the equation $$ ...


0

The eigenvectors tell you the direction of the flow locally around the fixed point. By locally, I mean, very close (how close? as close as you can be) to the fixed point. Here's a picture: The pure green and pure blue arrows represent the eigenvectors. The green eigenvector is associated with the negative eigenvalue, and the blue eigenvector is ...


0

suppose $$Lf = R,\quad f(0) = 0,\quad f'(0) = 1.$$ look at $$y=\int_c^xf(x-t)R(t)\,dt$$ then by leibniz rule of differentiaton, we have $$\begin{align}y' &= \int_c^xf'(x-t)R(t)dt + f(0)R(x)\\ &= \int_c^xf'(x-t)R(t)dt\\ y'' &= \int_c^xf''(x-t)R(t)dt+f'(0)R(x) \\ &= \int_c^xf''(x-t)R(t)dt+R(x) \\ Ly &= \int_c^x\left(af''(x-t) + bf'(x-t) ...


0

Property $2$ shows that $v$ is bounded on $[s,T)$. Since $b$ is continuous, the right hand side of $$ \frac{dv}{dt}=b(v(t)+f(t)) $$ is bounded on $[s,T)$. This gives $(*)$.


1

Note $t=(t-2)+2$. Then, $$\begin{align} \mathscr{L}\{tu(t-2)\}&=\mathscr{L}\{(t-2+2)u(t-2)\}\\ &=\mathscr{L}\{(t-2)u(t-2)\}+2\mathscr{L}\{u(t-2)\} \end{align}$$ Can you finish?


1

Going on the question you asked MV in his comments, I'll do it for you by separation of variables (note that using the Laplace Transform method MV used is a much better choice for this problem). We have $$U' = b_{n} - \bigg( \frac{n \pi}{L} \bigg)^{2} U$$ We will call $$m = \frac{n \pi}{L}$$ Separating and integrating $$\begin{align} \implies \int ...


0

firstly we should find the complementary solution $$y''+4y=0$$ $$m^2+4=0$$ $$m=\pm i$$ $$y_c=C1\cos(2t)+C_2\sin(2t)$$ the particular solution is $$y_p=(At+B)\cos(2t)+(Ct+D)\sin(2t)$$ when we compare this solution with complementary solution, we will find that there is similarity in two solution, so the particular solution should be ...


0

It'd be more accurate to write $$ (x-1)\frac{dy}{dx} = y^2 + 5$$ $$\frac{dy}{dx} = \frac{y^2+5}{x-1}$$ Since the RHS is a product of functions of $x$ and $y$, the equation is separable


2

If I understand your question, simply derive $y'$ and $y''$ from $y=e^{2it} v$ and plug in the above equation and then equate the coefficient of $e^{2it} $ in both sides. Note that $v$ is a function of $t$.


-4

make the ansatz $y_p=(At+B)\cos(2t)+(Ct+E)\sin(2t)$ with real numbers $A,B,C,E$


4

After setting $u=y/x$, the differential equation $$ \frac{dy}{dx}=\frac{2x-y}{x+2y} $$ becomes $$ x\frac{du}{dx}+u=\frac{2-u}{1+2u}, $$ i.e. $$ x\frac{du}{dx}=\frac{2-u}{1+2u}-u=\frac{2-2u-2u^2}{1+2u}=-\frac{2u^2+2u-2}{2u+1}. $$ Assuming that $$ 2u^2+2u-2\ne 0 \, \mbox{ and }\, x\ne 0, $$ we get $$\tag{1} \frac{2u+1}{2u^2+2u-2}du=-\frac1xdx. $$ Integrating ...


2

Hint: we have $$\frac{2-u}{1+2u}-u=\frac{2-u-u(1+2u)}{1+2u}=\frac{2-2u-2u^2}{1+2u}$$


0

The central difference formula for second derivative is $$f''(x_i)=\frac{f(x_{i+1})-2f(x_i)+f(x_{i-1})}{h^2}+O(h^2)$$ The central difference formula for first derivative is $$f'(x_i)=\frac{f(x_{i+1})-f(x_{i-1})}{2h}+O(h^2)$$ Both of these formulas are second order. Since the error terms are $O(h^2)$. So for an interior point $x_i$, plugging these into ...


2

Let's try the third power \begin{align} 2xf(x)^3&=e^x\{f(x)^2+\tfrac12 (f(x)^2)'\}\quad|·2x,\quad\text{insert original equation}\\ 4x^2f(x)^3&=e^x\left\{\left[e^x(f(x)+f'(x))\right]+x\left[e^x\frac{f(x)+2f'(x)+f''(x)}{2x}-e^x\frac{f(x)+f'(x)}{2x^2}\right]\right\} \\ 8x^3f(x)^3&=e^{2x}\left\{(3x-1)f(x)+(4x-1)f'(x)+xf''(x)\right\} \end{align} ...


1

Let $v = \frac{dx}{dt}$, then $$\frac{d^2x}{dt^2} = \frac{dv}{dt} = \frac{dx}{dt}\frac{dv}{dx} = \frac{v}{\frac{dx}{dv}}$$ Equation becomes $$ \frac{v}{\frac{dx}{dv}} + v + x = 5 $$ Differentiating with respect to $v$ $$\frac{1}{\frac{dx}{dv}} - \frac{v}{\left(\frac{dx}{dv} \right)^2}\frac{d^2x}{dv^2} + 1 + \frac{dx}{dv} = 0$$ $$ ...


1

$x$ depends on $y$. So : $$\frac{d}{dy} \left( \frac{x}{y} \right)=\frac{d}{dy} \left( x \cdot \frac{1}{y} \right)=\frac{dx}{dy} \cdot \frac{1}{y}+x \frac{d}{dy}\left(\frac{1}{y} \right)=\frac{dx}{dy} \cdot \frac{1}{y}+x \left(-\frac{1}{y^2} \right)$$


2

This is really just the quotient rule: $$\frac{d}{dy} \left ( \frac{x}{y} \right ) = \frac{y \frac{dx}{dy}-x}{y^2} = \frac{1}{y} \frac{dx}{dy} - \frac{x}{y^2}.$$ Here $\frac{dx}{dy}$ is the derivative of $x$ with respect to $y$, while $\frac{d}{dy} \left ( \frac{x}{y} \right )$ is the derivative of $\frac{x}{y}$ with respect to $y$.


1

Starting from $$U_n'(t)+\left(\frac{n\pi}{L}\right)^2 U_n(t)=b_n$$ with $U_n(0)=a_n$, we can find a solution for $U_n(t)$ using a variety of methods. Here, we will use LaPlace Transforms. Let $\hat U_n(s)$ be the Laplace transform of $U_n(t)$. Taking the Laplace transform of the ODE reveals $$(s\hat U_n(s)-U_n(0))+\left(\frac{n\pi}{L}\right)^2 \hat ...


1

For that you would need an actual Lagrange functional. Something like $$ L[x]=\int_a^b\left(\frac12 \dot x(t)^2-V(x(t))\right)\,dt $$ Then you apply it to a perturbed curve $x+δx$, or more precisely to a family $x+s·δx$ of perturbed curves, and compute the part linear in $δx$ $$ δL[x;δx]=\lim_{s\to0}\frac{L[x+δx]-L[x]}s = \int_a^b\left(\dot x(t)·δ\dot ...


1

Note that, under suitable conditions, the chain rule applies with $\frac {df}{dt}=\frac {df}{dy}\cdot \frac {dy}{dt}$ Use this with $f=x, x', x''$ and $y=x'$.


1

Set $z(t)=e^{-3t}y(t)$ and your equation reads $$z'(t)=C.$$ Integrating, you have $$z(t)=Ct+C',$$ i.e. $$e^{-3t}y(t)=Ct+C',$$ thus $$y(t)=e^{3t}(Ct+C').$$


1

You need to use the product formula for derivatives: $(f(t)g(t))'=f'(t)g(t)+f(t)g'(t)$. In your case : $f(t)=e^{-3t}$ and $g(t)=y=y(t)$. EDIT: Following our discussion in the comments, here is more details. We compute the derivatives of $f$ and $g$ as above: $$f'(t)=\frac{d}{dt}(e^{-3t})=e^{-3t}\frac{d}{dt}(-3t)=e^{-3t}(-3)=-3e^{-3t}$$ and ...


0

I would comment on the original post, but I don't have the rep yet. As is, the problem does not seem correctly posed. Consider the initial value problem $$y'=y^{1/3}, \hspace{1cm} y(0)=1.$$ Here, $f(x)=1$ and $g(y)=y^{1/3}$. $g$ is odd and strictly increasing. As $g$ is differentiable on the positive real line, it is Lipschitz continuous in any finite ...


0

It really depends what source you're working with. A lot of elementary differential equations texts will say that the general form for a second-order ODE is the more general $$P(x)u''(x)+Q(x)u'(x)+R(x)u(x)=g(x)$$ that you quoted. The right question to answer is: does it make sense to talk about solutions of a differential equation such as ...


2

the characteristic equation is $m^2+4=0$ $m=\pm2 i$ $$y_c=C_1\cos2 t+C_2\sin2 t$$ $$y_p=At^2+Bt+C+De^t$$ $$y'_p=2At+B+De^t$$ $$y''_p=2A+De^t$$ substitute in the original equation $$2A+De^t+4At^2+4Bt+4C+4De^t=t^2+7e^t$$ so $$A=1/4$$ $$B=0$$ $$C=-1/8$$ $$D=7/5$$ the general solution is $$y=C_1\cos2 t+C_2\sin2 t+t^2/4-1/8+7/5e^t$$


1

The characteristic equation is $r^2 + 4 = 0$, so you get $y_h = c_1 \cos (2 t)+ c_2 \sin(2 t) $. Now, use the method of undetermined coefficients to make some guesses to the general solution: If you plug in $y_3 = c_3 e^t$, you'll get $y_3''+ 4 y_3 = 5 c_3 e^t$, so having $c_3 = \frac{7}{5}$ gives the $e^t$ on the right. Now, for the $t^2$, since you have a ...


2

Here are a few perspectives; they reflect my opinion, so they may be coloured slightly by my exposure to mathematical biology. 1) $R_0$ is thought of as an expected number of secondary infections caused by a typical infectious individual during its infectious lifetime. This definition makes no reference to the size of the total population, and being a ...


0

Regarding (2), it is not true that if $f=\widehat{g}$, for some $g\in L^{1}(\mathbb{R})$, then $$\lim_{\epsilon\downarrow 0}\int_{\left|x\right|\geq\epsilon}\dfrac{f(x)}{x}dx$$ exists. To see this, consider $g=\chi_{[0,1]}$ (i.e. the characteristic function of the unit interval). Then $$\widehat{g}(\xi)=\int_{0}^{1}e^{-2\pi i y\xi}dy=-\dfrac{e^{-2\pi ...


1

Faced with $$ P(x)u''(x) + Q(x)u'(x) + R(x)u(x) = g(x) $$ if $P(x_0) \ne 0$ I can divide: $$ u''(x) + \frac{Q(x)}{P(x)}u'(x) + \frac{R(x)}{P(x)} u(x) = \frac{g(x)}{P(x)} \\ u''(x) + \hat{Q}(x)u'(x) + \hat{R}(x) u(x) = \hat{g}(x) $$ If $P(x_0) = 0$ I would analyze the order-reduced equation instead $$ Q(x)u'(x) + R(x)u(x) = g(x) $$ and again either get a ...


0

Here we have that $\frac{\partial}{\partial t}(f(x+ct))=\frac{\partial}{\partial t}(x+ct)f'(x+ct)=cf'(x+ct)$, this obtained by the composite function differentiation rule $(f\circ g)'=(f'\circ g)g'$. You do not need a substitution, the fact that $f$ is differentiable on all of $\Bbb R$ is enough.


0

I would just plug in the solutions to see what the equation looks like $$ g_1'' + ag_1' + bg_1 = (b-1)\sin x + a\cos x + (a+b+1)e^x = f(x) $$ $$ g_2'' + ag_2' + bg_2 = (b-1)\sin x + a\cos x + (a-b-1)e^{-x} = f(x) $$ This means $$ \begin{align} f(x) &= (b-1)\sin x + a\cos x + (a+b+1)e^x \\&= (b-1)\sin x + a\cos x + (a-b-1)e^{-x} \end{align}$$ Since ...


2

I'm assuming you know the method of characteristics, so we want the solution $\xi$ to be constant in the direction $(1,-p_1)$ (obtain by looking at the coefficients of $\partial_x\xi$ and $\partial_y\xi$). So we want the derivative in the direction $(1,-p_1)$ to be zero, i.e., $$\nabla \xi\cdot(1,-p_1)=0.$$ (remember ...


4

I'm assuming you want asymptotics for large $x$. Changing variables $x \to (x+1)$ your equation can be written more simply as $f'(x) =f( x- x^t +1)$ Let's write $f(x) = e^{g(x)}$. Then: $f'(x) = g'(x) e^{g(x)}$. $f(x - x^t +1 ) = e^{ g(x)} e^{ g(x-x^t+1) - g(x)} \approx e^{g(x)} e^{ - (x^t-1) g'(x) } $ This approximation is accurate if $g''(t)$ is not ...


1

derivative of $$y=x'+2x-e^{-3t}-e^{-t}$$ is $$y'=x''+2x'+3e^{-3t}+e^{-t}$$ substitute in the second equation to get $$x''+2x'+3e^{-3t}+e^{-t}+2(x'+2x-e^{-3t}-e^{-t})=e^{-3t}-e^{-t}+x$$ $$x''+4x'+3x=0$$ $$m^2+4m+3=0$$ $$m_1=-1$$ $$m_2=-3$$ so $$x=C_1e^{-t}+C_2e^{-3t}$$ $$x'=-C_1e^{-t}-3C_2e^{-3t}$$ so ...


1

Put them in matrix form: let $X(t)=\left( \begin{array} {c} x(t) \\ y(t) \end{array}\right)$; then $\left( \begin{array} {cc} 3 & 1 \\ 1 & 4 \end{array}\right) \dot X (t)+ \left( \begin{array} {cc} 5 & -1 \\ -2 & 7 \end{array}\right) X (t) = \left( \begin{array} {cc} 2 & 4 \\ -3 & 5 \end{array}\right) \left( \begin{array} {c} ...


1

You have that $(1,-1)\cdot\nabla u(x,t)=0$. This means that $u$ is constant on lines parallel to $(1,-1)$. Since the line connecting $(x,t)$ to $(x+t,0)$ is parallel to $(1,-1)$ we know that $$ u(x,t)=u(x+t,0)=f(x+t) $$ Check out The Method of Characteristics.


2

The point is that the $a_i$ and $g$ are fixed, known functions of $t$: the unknown function $y$ only appears in a linear combination of it and its derivatives.


0

The Picard-Lindelöf theorem helps in such cases. However, it is easy to construct solutions which do not have a unique solution, i.e. the theorem is not applicable. Consider for example $y^\prime(x) = x^{\frac{1}{3}}$ where $y(0)=0$. Then a solution is $y_1(x)=0$ and two others are $y_{2,3}(x) = \pm \left(\frac 23 x\right)^{\frac 32}$.


0

The standard Picard-Lindelöf-Cauchy-Lipschitz existence and uniqueness theorem will apply as long as $f$ is continuous and $g$ is locally Lipschitz.


0

Just compute $y''-y$ for your given solutions. I.e., you certainly want $g_1''(x)-g_1(x)=f(x)$ to hold.


2

You have $$e^{-y(s)}y'(s)=1,$$ hence $$\left(e^{ -y(s )}\right)'=-1.$$ Integrate with respect to $s$ from $0$ to $x$: $$e^{-y(x)}-e^{-y(0)}=0-x,$$so $$y(x) =-\ln(1-x).$$ This should be sufficient to find the parameter $a$.


2

Typically differential equations from physical modeling come equipped with parameters which have units. The most basic case is when $x$ has some unit (length, population, ...) and $t$ has the usual units of time. If you rescale $x$ and $t$ in an appropriate way, you can cancel out some of the parameters, in the process making the actual parameters in the new ...


1

It's easier than you think. $$x(t)= A\sin(t)+B\cos(t)$$ Now differentiate to find $x'(t)$ $$x'(t)= A\cos(t) - B\sin(t) $$ Sub this into your original equation and solve for A and B $$A\cos(t) - B\sin(t) - 3A\sin(t) - 3B\cos(t) = \frac{\cos(t)}{2}$$ Simplifying to $$\cos(t)[A - 3B - \frac12] - \sin(t)[B + 3A] = 0$$ Since there is no $\sin(t)$ or ...


1

We have $y''(t)=g(t,y(t))$. If we integrate both sides of this, we obtain $$y'(t)=y'(0)+\int_0^t g(t',y(t'))dt'$$ whereupon a second integration reveals $$\begin{align} y(t)&=y'(0)t+y(0)+\int_0^t \int_0^{t''} g(t',y(t'))dt'dt''\\\\ &=y'(0)t+y(0)+\int_0^t \int_{t'}^{t} g(t',y(t'))dt''dt'\\\\ &=y'(0)t+y(0)+\int_0^t g(t',y(t'))\int_{t'}^{t} ...


-1

It is easy to answer your second question with Maple: dsolve(((D@@2)(w))(z)+sin(z)*(D(w))(z)+(z^2+1)*w(z) = 0, w(z), series, order = 10); $$ w \left( z \right) =w \left( 0 \right) +\mbox {D} \left( w \right) \left( 0 \right) z-{\frac {w \left( 0 \right) }{2}}{z}^{2}-{\frac { \mbox {D} \left( w \right) \left( 0 \right) }{3}}{z}^{3}+{\frac {w \left( 0 ...



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