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1

$$\frac{dy}{dt}= aty +4e^{-t^2}$$ An explicit formula for the solutions is : $$y=c e^{\frac{a t^2}{2}}+2\sqrt{\frac{2\pi}{a+2}}e^{\frac{a t^2}{2}}erf \left(\sqrt{1+\frac{a}{2}}\space t\right)$$ but I suppose that the function erf is not allowed as well an integral. The solutions of the homogeneous ODE $\frac{dy}{dt}= aty $ are : $$y=C e^{\frac{a t^2}{2}}$$ ...


1

Your suspicions are correct, you can convert an $n-th$ order differential equation into an n-dimensional system of first order equations. Let $x_1 = x$, and then we have: $$\begin{align} x_1' &= x' = x_2 \\ x_2' &= x'' = x_3 \\ x_3' &= x''' = -t^2 x_3 - 4 x_1 \end{align}$$ Our new system is: $$\begin{align} x_1' &= x_2 \\ x_2' &= ...


2

$\mathcal{L}(y)$ is correctly computed. Your error is in computing the inverse Laplace transform: $$ \frac{69-8\,s}{(s-4)^2+11}=\frac{37}{(s-4)^2+11}-\frac{8(s-4)}{(s-4)^2+11}. $$


0

This is one of the rare occasions where it actually helps to differentiate the given ODE! In this way we obtain $$2r'(t)\bigl(r(t)+r''(t)\bigr)=0\qquad\forall t\ .\tag{1}$$ This is satisfied when $r'(t)\equiv0$, or $r(t)=c$ for some $c\in{\mathbb R}$. From the original ODE it then follows that $c^2=1$, so that we obtain the two solutions ...


1

I doubt this has an elementary analytical solution for general $\gamma$, but we can easily construct solutions by taking any function $g(t)$ with $g(0)=0$ and $g'(0)=1$ and then take $\gamma = gg''$. By construction we have that $f(t)=g(t)$ is a solution. One case we can solve is $\gamma(t) = C$ since then $$f'' f = C \implies f''f' = \frac{Cf'}{f} ...


0

Your equation is only of first-order, only first derivatives with respect t only one variable appear there. There is no classification of such equations into elliptic, parabolic, hyperbolic equations. If you talk about systems of first-order equations, they can be classified, if they can be written as second-order equations. E.g. the system $$ ...


3

A Dirichlet condition is the prescription of the values of the solution at the boundary. A zero Dirichlet condition means the solution needs to be $0$ there. So you are asked to find the $\theta$ so that for this choice of $\theta$, you have $u(0,t)= 0$ and $u(l,t)=0$. Practically speaking, you set $u(0,t)= 0$, with the function $u$ you have, and solve ...


1

If you compare two operation: differentiation and integration, then the clear winner is the integration, you can integrate way more functions than differentiate (moreover, integration is a numerically stable operation, whereas numerical differentiation is an ill posed problem). On the other hand, what is easier to do in Calc, differentiate or integrate? ...


1

I would recommend starting with the linear problem $\tilde{y}′(x)=\underbrace{\frac{8A2x}{(1+4A2x^2)^2}}_{=:M(x)}⋅\tilde{y}(t)$ using a separation $ \frac{d\tilde{y}}{\tilde{y}} = M(x) dx $ to get $\tilde{y} = e^{c + \int M(x) dx}$ From this solution you can use basic techniques to get an answer for the actual problem.


1

To check fixpoint stability we linearize the equations around the fixpoint $(x,y) = (\overline{x} ,\overline{y})$. Put $x= \overline{x} + \delta x$, $y= \overline{y} + \delta y$ and expand the equations to first order in the perturbations. For a general system $$\dot{x} = F(x,y)$$ $$\dot{y} = G(x,y)$$ the fixpoints are determined by ...


2

To solve the equation rewrite it as $$r^2(u_{rr} + \frac{2}{r}u_r + u) = r\left[\frac{d^2}{dr^2}\left(ru\right) + (ru)\right] = 0$$ leaving us to solve $y_{rr} + y = 0$ where $y = ur$. This equation has the general solution $y = A\sin(r) + B\cos(r)$. Regularity at $r=0$ (this condition is usually imposed) implies $B=0$ and we are left with $$u(r) = ...


0

the first part is correct(can be solved by using the characteristic polynomial), to find the constants, you must derive the initial equation and plug in the data(i.e. putting 0 in and so that y(0)=2, and like so in y'(0)=0), this will give you 2 equations with 2 variables. so you can easily find the constants. As for the 2 cosh 2t, this seems to be a ...


1

Just find the values and plug them in: $$\begin{cases}y=c_1e^{2t}+c_2e^{-2t}\\y'=2c_1e^{2t}-2c_2e^{-2t}\\y''=4c_1e^{2t}+4c_2e^{-2t}\end{cases}\\y''-4y=4c_1e^{2t}+4c_2e^{-2t}-4(c_1e^{2t}+c_2e^{-2t})=0$$ Second part you're given initial conditions: $$y(0)=2=c_1e^{2(0)}+c_2e^{-2(0)}=c_1+c_2\\y'(0)=0=2c_1e^{2(0)}-2c_2e^{-2(0)}=2c_1-2c_2$$ So you get: ...


0

Simply substitute $t-0$ into the expression for $y(t)$ to get:$$0=C_1e^0+C_2e^0=C_1+C_2\tag{1}$$then differentiate $y(t)$ to get:$$y'(t)=2C_1e^{2t}-2C_2e^{-2t}$$and again substitute $t=0$ into this to get:$$0=2C_1-2C_2\tag{2}$$Now you have two equations with two unknowns which you can solve to find $C_1$ and $C_2$.


1

The heat equation is parabolic: $$ \frac{\partial f}{\partial t} = \frac{\partial^{2}f}{\partial^{2}x}+\frac{\partial^{2}f}{\partial y^{2}}+\frac{\partial^{2}f}{\partial z^{2}} $$ Laplace's equation is elliptic: $$ \frac{\partial^{2}f}{\partial^{2}x}+\frac{\partial^{2}f}{\partial y^{2}}+\frac{\partial^{2}f}{\partial z^{2}} = 0 $$ The ...


1

Of course not. Actually most differential equations can't be solved neither by quadratures (Google Liouville theory analogue of Galois theory) nor by other methods. It is also very easy to produce an equation for which we can't describe phase space. The only partially complete theory exists for Linear differential equations and more particularly for Linear ...


0

Hint: Check Riccati differential equation.


2

this method of reducing the order of a linear differential equation is called the variation of parameter. what you do is assume you have found one solution $y$ of $$x^{\prime \prime} + px^\prime + qx = 0 \tag 1$$ now you assume a solution of the form $$x = yu, x^\prime = y^\prime u + yu^\prime, x^{\prime \prime}= y^{\prime \prime}u+2y^\prime u^\prime +y ...


1

Hint: Employ the identity $$\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B),$$ within $c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x)$.


0

$3hx^2-2x+5hx=3\implies$ $3hx^2+(5h-2)x-3=0\implies$ $x_{1,2}=\frac{2-5h\pm\sqrt{(5h-2)^2+4\cdot3h\cdot3}}{2\cdot3h}\implies$ $x_{1,2}=\frac{2-5h\pm\sqrt{25h^2+4h+4}}{6h}$ $x_1+x_2=-3\implies$ $\frac{2-5h+\sqrt{25h^2+4h+4}}{6h}+\frac{2-5h-\sqrt{25h^2+4h+4}}{6h}=-3\implies$ $\frac{2-5h+2-5h}{6h}=-3\implies$ $4-10h=-18h\implies$ $8h=-4\implies$ ...


0

First rewrite your equation in standard form as follows:$$3hx^2-2x+5xh=3$$$$\therefore3hx^2+(5h-2)x-3=0$$Assuming $h\ne0$ we can divide it all by $3h$ to get:$$x^2+\frac{(5h-2)}{3h}x-\frac{1}{h}=0\tag{1}$$Now, if a quadratic equation has roots $r_1$ and $r_2$, then it can be written as:$$(x-r_1)(x-r_2)=0$$$$\therefore x^2-(r_1+r_2)x+r_1r_2=0\tag{2}$$You are ...


0

A homogeneous differential equation have same power of $X$ and $Y$ example :$- x+y dy/dx= 2y$ $X+y$ have power $1$ and $2y$ have power $1$ so it is an homogeneous equation.


1

For part a), the problem asks you to draw the direction field without solving the ODE. Writing it in this form: $$y'=2e^{-t}-(1+\frac{1}{t})y$$ Try some $y$ values. For example, let $y=0,y'=2e^{-t}$. That will give you the direction field on $t$ axis. They are arrows pointing upward, but approaching 0 direction, toward some equilibrium value. Let $y=1, ...


0

Why is $-\ln(x)$ is not equal to $\frac{1}{\ln(x)}$ ? Because $\space -\ln(x)=\ln(\frac{1}{x})\space $ and $\space \ln(\frac{1}{x})\space $ is not equal to $\space \frac{1}{\ln(x)}$ In general, for most of the functions $f(x)$ we don't have $f(\frac{1}{x})=\frac{1}{f(x)}$


0

Where is no closed form for the solutions of the non linear ODE : $$ \frac{d^2y}{dx^2}+(\frac{1}{y})\frac{dy}{dx}=-1 $$ Of course, it can be solved thanks to numerical methods, but it is not what is expected. In your additional comment, you wrote : " Answer is $(1−x^2 )/4$ .....but i dont know how to solve ..." It is easy to check that $(1−x^2 )/4$ is not ...


1

Your equations amount to a system of $8$ scalar equations with $8$ scalar unknowns. Writing $x$ and $y$ as separate vectors is not helping clarity here. Stack them into one $8$-dimensional vector $z$, and similarly stack $f$ and $g$ into one function $h$. Formally speaking, the equation $h(z)=0$ could be solved by the Newton-Raphson method as $$ z_{n+1} = ...


1

As abel suggests, since $|\mathbf{v}(t)|^{2}$ is conserved, then this suggests spherical coordinates (as $r$ will be constant). Using the convention $x(t) = r\cos(\theta)\sin(\phi)$, $y = r\sin(\theta)\sin(\phi)$, $z = r\cos(\phi)$, we have \begin{align} \dot{x}(t) &= -r\sin(\theta)\sin(\phi)\dot{\theta} + r\cos(\theta)\cos(\phi)\dot{\phi} \\ \dot{y}(t) ...


1

Let me risk then the following Answer: The right interpretation of the case is that $$dw=\frac{kdt}{w(t)}.$$ And a solution for $k=1$ is $$w(t)=\sqrt{2t}$$ if $t \ge0$ and undefined otherwise. (Also, by common sense: $w(0)=0$.)


1

Here's a Python implementation of the midpoint method, which unfortunately doesn't exploit the skew-symmetric nature of the operator except as an assertion that $\left|\mathbf{v}(t)\right|^{2}$ is constant. I've been looking for an excuse to learn Matplotlib; this is somewhat of a minimal working example: #!/usr/bin/env python3 """This script parses user ...


0

I would do it like this. First let the whole distance be $12$ units (you can choose the units to fit and $12$ makes everything an integer). Now at some time $t_1$, Alison has travelled $9$ units so that $9=\frac {at_1^2}2$ and $$t_1^2=\frac {18}a$$ At time $t_2$, Alison has travelled $12$ units, so $$t_2^2=\frac {24}a$$ Now this last quarter of the ...


0

We have the general equation $s(t) = \frac{1}{2} at^2$, which will we can apply to any time in the race. If we let $S$ be the end of the race and $T$ be the time it takes for Kevin to reach the end, we will have $$ T = \sqrt{\frac{2S}{a}} $$ But, we also know that 3 seconds earlier, he was 1/4 of the total distance away from the finish line, or in other ...


0

After reviewing the calculations and talking to a colleague, it was a distraction of my part. The solution to the ODE is actually $$y = c_1e^x + c_2e^{-x} + xe^x,$$ I had solved it correctly before, but I don't know why I forgot that $x$ after. Then the system becomes $$\begin{cases} c_1 + c_2 = 0 \\ ec_1 + \frac{c_2}{e} + e = 1 \end{cases}.$$ (note my ...


0

The boundary conditions for this type of problem can often be formulated as a matrix equation yielding a fairly general and automatic way to find the eigenvalues. Here's how to do so for this problem. First, as you notice, the general solution of the differential equation is $$u(x)=a\cos(\sqrt\lambda x)+b\sin(\sqrt\lambda x)$$ so that ...


1

Knowing that $c_1J_2+c_2Y_2$ is the solution of the Bessel equation of the first kind : $$x^2y''+x y'+(x^2-4)y=0$$ The solution of $$x^2y''+x y'+(x^2-4)y=f(x)$$ is on the form $y=c_1J_2+c_2Y_2+y_p(x)$ where $y_p(x)$ is a particular solution. In the present case, $y_p(x)=c_3/x^2$. Hense : $$x^2y_p''+x y_p'+(x^2-4)y_p=f(x)$$ Now, you can find $f(x)$ I change ...


1

Here is my own try: $$\frac{\lambda y'}{\sqrt{1+y'^2}}=x+C$$ $$\frac{ y'}{\sqrt{1+y'^2}}=\frac{x+C}{\lambda}$$ $$\frac{y'^2}{1+y'^2}=\frac{\left(x+C\right)^2}{\lambda^2}$$ $$\frac{-1}{1+y'^2}=\frac{\left(x+C\right)^2-\lambda^2}{\lambda^2}$$ $$1+y'^2=\frac{-\lambda^2}{\left(x+C\right)^2-\lambda^2}$$ ...


1

hint: integrating once you get $$\dfrac{y^\prime}{\sqrt{1+{y^\prime}^2}} = \dfrac{x + C}{\lambda} \tag 1$$ (a) square $(1)$ and find $y^\prime$ (b) solve the first order equation. edit: since the left hand side of $(1)$ is smaller than one in absolute value, it can be set to $\sin \theta.$ which gives $$x + C = \sin \theta, \ ...


0

to make sense of the definition of order of a differential equation, you need to first isolate the highest order derivative first. for example, $\frac{dy}{dx} = y, \frac{dy}{dx} = y^2 + 1$ are of the first order. $ \frac{d^2 y}{dx^2} = y, \frac{d^2 y}{dx^2} = (\frac{dy}{dx})^3 + 1$ are of second prder. your differential equation $\frac{dy}{dx} = ...


0

No need for substitution really $$ f^2 + xff' = cf' \implies f + xf'-\frac{c}{f}f' = 0 $$ Now you can see $$ \left(xf\right)' -c\left(\ln f\right)' = 0 $$ You can integrate straight away to yield $$ xf -c\ln f + k = 0\tag{*} $$ Using the sub $ \ln u = \phi $ where $u^{-1} = f\mathrm{e}^{-k/c}$ (skipping a few steps in logic) we find $$ ...


2

Your mistake is in this line: $$ y'(x) = -5C_1e^{-5x} -5C_2xe^{-5x} \color{red}{+C_2 e^{-5x}} $$ by the product rule.


1

You have \begin{align} 3t^2\frac{dy}{dt}+t^3y=\cos\left(t\right).\tag{1} \end{align} Now put the equation in standard form as follows: \begin{align} 3t^2\frac{dy}{dt}+t^3y=\cos\left(t\right)\Leftrightarrow \frac{dy}{dt}+\left(\frac{t}{3}\right)y=\frac{\cos\left(t\right)}{3t^2}\tag{2}. \end{align} Notice that the LHS looks a lot like the product rule of ...


0

Yes and no. There isn't a single field that covers all of it. The closets analogy is probably the study of Dynamical Systems. In general, we can't find exact solutions to systems of linear DEs. However, for a certain type, called linear dynamical systems, we can solve them systematically. In fact, many of the tools you might have seen in linear algebra ...


1

$$\begin{align} \bigg(\frac{d^{4}}{dt^{4}} - \lambda \bigg) y &= \bigg( \frac{d^{2}}{dt^{2}} + \sqrt{\lambda} \bigg) \bigg(\frac{d^{2}}{dt^{2}} - \sqrt{\lambda} \bigg) y\\ &= \bigg(\frac{d}{dt} + i \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - i\lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} + \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - ...


0

You're right, you don't have to write it in the Sturm-Liouville form to find its eigenfunctions. But in that case you've got nothing more than that, and we wish to find useful properties. So would like to use this thereome that says the following. If you have an equation of the type: $$(py')'-qy=-\lambda w y$$ One usually assumes some regularity on $p, ...


1

let us take a simple operator $L = \frac{d}{dt} + y$ and look at the equation $$L y = \frac{dy}{dt} - y^2 = 0 \tag 1$$ we can verify that $y_1 = \dfrac{1}{1-t}$ and $y_2 = \dfrac{2}{2-t}$ are solutions of $(1)$ and $y_2(0) = 2y_1(0).$ if $L$ were linear we would have $y_2(t) = 2y_1(t)$ at least on the interval common existence. do we have that?


0

Here we compute the integral over the Bessel functions. Note that this integral can be written as: \begin{equation} {\mathcal I} = \int\limits_{t_0}^t \left| \begin{array}{cc} f_+(\eta) & f_-(\eta) \\ f_+(\xi) & f_-(\xi) \end{array} \right| \frac{\omega^2 \eta^{-\gamma-1}}{{\mathcal W}[f_+,f_-](\eta)} d \eta \end{equation} where \begin{equation} ...


3

We can factor out the variable dependence in this particular case: $$ A=\begin{pmatrix}e^{-2t} & e^{-t} & 0 \\ -\frac{5}{4}e^{-2t} & -\frac{4}{3}e^{-t} & e^{2t} \\ -\frac{7}{4}e^{-2t} & -\frac{2}{3}e^{-t} & -e^{2t} \end{pmatrix} $$ $$ = \begin{pmatrix} 1 & 1 & 0 \\ -\frac{5}{4} & -\frac{4}{3} & 1 \\ -\frac{7}{4} ...


2

The function $u^-(x,y) = f(bx-ay)$ solves $au_x+bu_y=0$, while $u^+(x,y) = f(bx+ay)$ doesn't, for the same reason that $x^+=3$ solves $4x-12=0$, while $x^-=-3$ doesn't - it's just a matter of plugging in. In the equation $4x-12=0$, the symbol $x$ represents a number. Any solution to the equation is some number with the property that, if you plug that ...


0

As an example, let $g(x):= \frac{1}{x^2}-1$, in which case the integral could diverge to $+\infty$ (depending on how fast $\phi(x)$ moves away from zero and the value of $x$ in the limits. In general,since $g(x)>0$ on $(0,1)$, and $\phi(x)\in (0,1),\;\forall x$ the integral must take only positive values. Thus, it must be either positive finite, or ...


0

You have made mistakes in your implementation of the Runge-Kutta Method. Please see the differences. Yours: kt1 = thetadot(u); kt2 = thetadot(u) + 0.5 * h * kt1; kt3 = thetadot(u) + 0.5 * h * kt2; kt4 = thetadot(u) + h * kt3; thetanext = thetadot(u) + (h / 6) * (kt1 + 2 * kt2 + 2 * kt3 + kt4); Corrected: kt1=thetadot(u); kt2=thetadot(u + 0.5 * h * ...


2

The solution, written in implicit form, is $$\frac1{18}\big(\ln|y + 9| - \ln|y - 9|\big) = \frac{x^2}{36} + C$$ Using the properties of the logarithms rewrite it in a more manageable form: $$\frac1{18}\ln\left|\frac{y + 9}{y - 9}\right| = \frac{x^2}{36} + C\tag{1}$$ Now, first of all substitute $x = 0$ and $y = 81$ to find $C$: $$C ...



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