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2

$exp(-409t)*d[exp(409t)*y(t)]/dt=f(t)$, so $exp(409t)*f(t)=d[exp(409t)*f(t)]/dt$, integrating, we have that $y(t)=exp(-409t)*integral[exp(409t)*f(t)]$. in part b), see my answer in Prove that $\lim \limits_{x\to\infty}f(x)=0$ and see that $y'(t)$ tends to $0$ when $t$ tends to infinity.


2

$$ \left(f'' +\frac{f'}{x}\right)(1+f'^2)=f'^2f'' $$ multiply by $x$ $$ (xf''+f')(1+f'^2) = \left(\dfrac{d}{dx}xf'\right)(1+f'^2) = xf'^2f'' $$ then we have $$ \frac{1}{xf'}\left(\dfrac{d}{dx}xf'\right) = \frac{f'f''}{(1+f'^2)} $$ thus $$ \dfrac{d}{dx}\ln(xf') = \frac{1}{2}\dfrac{d}{dx}\ln(1+f'^2) $$ which is $$ \ln (xf') = \frac{1}{2}\ln(1+f'^2) + C $$ or ...


0

Let me first say that which fixed points are stable/unstable and then the reasons. From the equation $4y^2(4-y^2)$, the fixed points are $y=0$, $y=-2$, and $y=2$. The first one is inconclusive, it could be stable or unstable depending on where you start your trajectory. $y=-2$ is unstable and $y=2$ is stable. Now, there are two ways to investigate the ...


2

The system you propose is fine. You should find the following generalized eigenvectors: $$\vec{v}_3=\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\quad\vec{v}_4=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$$


2

First out, definition of Generalized Eigenvector, So you would want to try and solve $({\bf A}-\lambda {\bf I})^k{\bf v} = {\bf 0}$, but $({\bf A}-\lambda {\bf I})^{k-1}{\bf v} \neq {\bf 0}$ to answer your question. You can use the first order eigenvectors (ordinary eigenvectors) together with the block-zero property to limit the span of generalized ...


0

I would rewrite it to $$ y' = -2(y-\tfrac12 e^{-t})-y^3 $$ Intuitively the first term pulls $y$ towards $\frac12 e^{-t}$ and the second term pulls it towards $0$. So eventually it should be expected to end up between those targets (and thus converge towards $0$). More formally, if you have a starting condition, you can bound the solution from below by the ...


1

I suppose you have already obtained its solution by separation of variables. Perhaps needing to recognize special cases. $ f_2 = 0 $ describes standard spring mass system of second order ODE including damping effect.The time curves are three types. Over-damped, critical damped and under / oscillatory damped. Case $ f_1 = 0 $ is valid for motion in viscous ...


1

The exact solutions to Volterra-Lotka are convex, almost circular curves. The explicit method follows the tangents of these curves, which means that every step changes to a more outward curve. The implicit method is the exact reverse, each step changes the level to a more inward curve. You need higher-order explicit methods for better preservation of the ...


0

It is a separable equation: $$ \frac{m\,v'}{f_0+f_1\,v+f_2\,v^2}=1. $$ The solution is $$ \int_0^{v_0}\frac{m\,dv}{f_0+f_1\,v+f_2\,v^2}=t. $$


1

To solve for $x$ make the change $x=t\,v$. This will give an equation for $v$ that can be solved by separation of variables.


0

Hint: $\dfrac{\partial^2u}{\partial r^2}+\dfrac{1}{r}\dfrac{\partial u}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2u}{\partial\theta^2}=0$ Let $u(r,\theta)=G(r)\Theta(\theta)$ , Then $G''(r)\Theta(\theta)+\dfrac{G'(r)\Theta(\theta)}{r}+\dfrac{G(r)\Theta''(\theta)}{r^2}=0$ ...


0

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$ $\dfrac{dz}{dt}=2$ , letting $z(0)=z_0$ , we have $z=z_0+2t=z_0+2y$ $\dfrac{dx}{dt}=1+\sqrt{z-y-x}=1+\sqrt{z_0+t-x}$ Let $u=z_0+t-x$ , Then $x=z_0+t-u$ $\dfrac{dx}{dt}=1-\dfrac{du}{dt}$ ...


0

I would recommend: Canuto C. et al. Spectral Methods (springer link). Trefethen L. Spectral Methods in Matlab. (This last one guides you through the implementation of the codes in Matlab but it can be easily extended to other programming language). Hope it helps. Cheers!


1

When you substitute into the equation you have: $2Ae^t + Ate^t -3(Ae^t + Ate^t) +2(A t e^t)=e^t \implies e^t(-3At+3At+2A-3A)=e^t $ So A=-1


0

For the particular solution, try $y_p=t(c_1e^t)$ where the $t$ us given to account for the $e^t$ in the complementary solution (I will go through that if you like).


1

You have changed the right-hand side of the equation to $te^t$ instead of $e^t$.


1

Let's substitute $x'=y$ and solve $$y'+y=5t\cos{t}+4\sin{t}$$ The homogeneous equation has got a general solution of the form $y(t)=Ce^{-t}$. Let's look for a solution of the equation above of the form $y(t)=C(t)e^{-t}$. substituting we get $$C'(t)e^{-t}=5t\cos{t}+4\sin{t}$$ So $$\begin{align} C(t)&=\int e^{t}\left(5t\cos{t}+4\sin{t}\right)dt\\ ...


1

We can solve this using undetermined coefficients. The complementary solution is found by solving \begin{equation*} x''+x'=0. \end{equation*} Substituting $x=e^{\lambda t}$ and factoring out $e^{\lambda t}$ gives \begin{equation*} (\lambda^2+\lambda)e^{\lambda t}=0. \end{equation*} The zeros must come form the polynomial giving $\lambda =-1$ or $0$. The ...


1

is it not easier to solve the first order equation $$\dot y + y = 5t\cos t + 4\sin t, y = \dot x.\tag 1$$ try a particular solution of the from $$y = a\cos t + b \sin t + ct\sin t,\\ \dot y = -a \sin t + b \cos t+c\sin t +ct\cos t\tag 2$$ subbing $(2)$ in (1), you get $$ct\cos t +(a+b)\cos t +(-a+b)\sin t = 5t\cos t+4\sin t $$ equating the coefficients ...


0

It means that you differentiate with regard to x, and then divide by x, and then repeat the entire process $($by repeatedly applying these two operations, in this exact order$)$, a total number of n times.


1

We have $$y'=y_1y_2'+y_2y_1'=(y_1y_2)'$$ And so a solution is $y(x)=y_1(x)y_2(x)$


0

Note that you connect $a_n$ and $a_{n-1}$. In the index shift you had that right, in the recursion equation (lose the sum sign) this changed. Using the equation for degree $n=0$ and setting $a_1=0=a_{2m+1}$ or using the equation for degree $n=1$ and setting $a_0=0=a_{2m}$ results in the same solutions, since that index shift is compensated by the root ...


1

Using series, write $$x=\sum_{i=0}^\infty a_it^i$$ $$x'=\sum_{i=0}^\infty ia_it^{i-1}$$ $$x''=\sum_{i=0}^\infty i(i-1)a_it^{i-2}$$ Rewrite $$t^2x''-(6t^4+2t)x'+9t^6x=t^2x''-6t^4x'-2tx'+9t^6x$$which gives $$A=\sum_{i=0}^\infty i(i-1)a_it^{i}-6\sum_{i=0}^\infty ia_it^{i+3}-2\sum_{i=0}^\infty ia_it^{i}+9\sum_{i=0}^\infty a_it^{i+6}=0$$ So, for given power $m$ ...


0

2(r-1)r+r reduces to the answer you are looking for. I dont see a problem with the indicial equation


0

Starting with your Euler equation \begin{align} x^{2} y'' + \alpha x y' + \beta y &= y'' + \frac{\alpha y'}{x} + \frac{\beta y}{x^{2}} \\ &= 0 \end{align} we can see that our ODE will be undefined at $x = 0$ unless $$\frac{\alpha}{x} \ \ \ (1)$$ and $$\frac{\beta}{x^{2}} \ \ (2)$$ are analytic at $x = 0$. For example, if $\alpha$ and $\beta$ ...


1

In the first step of your solution, you're only able to divide $y' = t^3y$ by $y$ if you assume that $y \neq 0$. You're excluding that solution.


0

Hint: Consider $X(x)=a\sin(\lambda(x-1))+b\cos(\lambda(x-1))$ $X(1)=0$ : $b=0$ $\therefore X(x)=a\sin(\lambda(x-1))$ $X'(x)=a\lambda\cos(\lambda(x-1))$ $X'(0)-X(0)=0$ : $a\lambda\cos(-\lambda)-a\sin(-\lambda)=0$ $\lambda\cos\lambda+\sin\lambda=0$ $\tan\lambda=-\lambda$ Hence for $\dfrac{\partial u}{\partial t}=\dfrac{\partial^2u}{\partial x^2}$ ...


0

I will analyze $x>6$ and $x=6$ is an irregular singular so we can use what is called a WKB Approximation. If you have an equation of the form: $$y''=Qy$$ then the asymptotic solution of that equation will be of the form: $$y=\frac{exp(\pm \int\sqrt Q d x)}{Q^{1/4}}$$ Lets try to approximate the equation with initial conditions $y(0)=1$ and $y'(0)=0$. ...


0

$y(x)=0$ and $y(x)=1$ are solution of the differential equation. By uniqueness these are the only solutions with the initial conditions $y(0)=0$ and $y(0)=1$ respectively. In your solution, in fact, you were supposing that $y\neq0,1$ everywhere.


0

I assume you mean $y_{0}=y(0)$ so that $x_{0}=0$. As you rightly point out, the initial value problem $y'=y(y-1)\\y(0)=0$ cannot be solved by separation of variables. The same problem arises if $y_{0}=1$ and the solution is similar: Since $F(x,y)=y(y-1)$ has a continuous partial derivative in $y$ you know there is a unique solution. Can you find it by ...


2

First of all, you should see that since $y=0$ is a root of the right side that $y\equiv 0$ is a steady state or constant solution, as is $y\equiv 1$. Secondly, you can transform your final expression to $$ y(x) = \frac{y_0}{y_0-(y_0 - 1)e^x}. $$ which is also valid for $y_0=0$.


0

There is a typo in the equation $\frac{d}{dx}((1-x^2)^\frac{1}{2}\frac{dy}{dx}) + n^2(1-x^2)^\frac{1}{2}y=0$ because the solutions are in contradiction with what is expected latter: The solutions involves the Mathieu's functions. Certainely, the equation is : $$\frac{d}{dx}((1-x^2)^\frac{1}{2}\frac{dy}{dx}) + n^2 \frac{1}{(1-x^2)^\frac{1}{2}}y=0$$


0

Letting capital letters denote the laplace transform and taking the laplace transform of both equations, we get $$ sX(s)-x(0) = 2X(s) + 3Y(s) $$ $$ sY(s)-y(0) = 3X(s) + 2Y(s) $$ Collecting like terms we get $$(s-2)X(s)-3Y(s)-x(0)=0$$ $$(s-2)Y(s)-3X(s)-y(0)=0$$ Using elimination, we can get $$-9Y(s)-3x(0)+(s-2)^2Y(s)-(s-2)y(0)=0$$ Moving things around a ...


1

Hint: Shift the bounds of the summation to to match the powers of $x$. You will have one starting at $n=0$ and one starting at $n=1$. Take out the $0 ^{th}$ term from the one that has the bound $n=0$ and combine the two summations which now have the same $x$ power. You will have a recursive relationship between $a_n$'s.


1

Waheguru ji ka khalsa, waheguru ji ki fateh. $$ D^n\left(x^n\right) = n!\\ D^n\left(x^{n-1}\right) = 0\\ $$ then we can find $$ y'' = 0\tag{1} $$ $$ y''' = 6.\tag{2} $$ we can solve the first one we know that the highest order of Eq. 1 is $n-1 = 2-1 = 1$ i.e. $$ y = x + c\ $$ for the second one we know that $3! = 6$ this means the highest order is $x^3$ so ...


0

Example 1. Consider equation number $2$. Since its third derivative is constantly zero, the solution must be a polynomial of degree $d<3$: otherwise, the derivative would not be constantly zero. Example 2. Consider equation number $5$. This time, its third derivative is constantly $6$. This has a correlation with the leading coefficient of the ...


1

Expanding on Sonnhard's hint, on the left hand side you have $$\Big(1+\frac{1}{u-1}\Big)du$$ Integrate this gives you $u+\text{ln}|u-1|+C$, where $C$ is some arbitrary constant.


2

HINT: we have $$\frac{du}{1-\frac{1}{u}}=\frac{dx}{x}$$ the term $$\frac{u}{u-1}=\frac{u-1+1}{u-1}=1+\frac{1}{u-1}$$


0

I do not understand how this is different but in general it is extremely useful to be able to recognize these things: $$y^2y'=\frac{(y^3)'}{3}$$ $$yy'=\frac{(y^2)'}{2}$$ $$xy'+y=(xy)'$$ These come up all the time.


2

Anyway here's a solution, since you are asking for intermediate steps. $3y^2dy+x^2dy+2xdx y+xdy +ydx +2ydy+dy=0$ $\color{blue}{\overbrace{3y^2dy+2ydy+dy}}+\color{red}{\underbrace{x^2dy +2xdxy}} + \color{green}{\overbrace{xdy + ydx}}=0$ Notice the expression coloured, They can be written as $\color{blue}{3y^2dy+2ydy+dy=d(y^3+y^2+y)}$ $\color{red}{x^2dy ...


0

The equation $$\frac {dy}{dx}=\sqrt {\frac {k-y}y}$$ is equivalent to $$\frac {dx}{dy}=\sqrt {\frac y{k-y}}$$ so you can find $x$ by an integration with respect to $y\,$. Use the substitution $$\sqrt {\frac y{k-y}}=\tan \phi$$ so $$y=k\, \frac {\tan^2 \phi}{1+\tan^2 \phi}=k\, \sin^2 \phi$$ Then $$x= \int \sqrt {\frac y{k-y}}\,dy= \int \tan \phi \cdot 2k ...


0

Define $$ w(x)=\frac{1}{2}(1-x_1^2) \max_{S}f^+, x=(x_1,x_2). $$ Here $s^+=s$ if $s\ge0$ and $s^+=0$ if $s<0$. Then, in $S$ $$ -\Delta (w-u)=\max_{s}f^++\Delta u=\max_{s}f^+-f\ge0, $$ and on $\partial S$, $w-u\ge0$. By the Maximum Principle, $w-u\ge 0$ in $S$ or $$ u\le w=\frac12(1-x_1^2) \max_{S}f^+\le\frac12\max_Sf^+. $$ Similarly we have, in $S$ ...


4

rewrite it in the form $$\frac{dy}{y}=\left(\frac{x+1}{x}\right)dx$$


1

$$xy'_{(x)}=y_{(x)}(x+1)\Longrightarrow$$ $$xy'=y(x+1)\Longleftrightarrow$$ $$y'=y(x+1)\frac{1}{x}\Longleftrightarrow$$ $$\frac{dy}{dx}=y(x+1)\frac{1}{x}\Longleftrightarrow$$ $$\frac{1}{y}dy=\frac{1}{x}(x+1)dx\Longleftrightarrow$$ $$\int\frac{1}{y}dy=\int \frac{1}{x}(x+1)dx\Longleftrightarrow$$ $$\ln|y|=\int (1+\frac{1}{x})dx\Longleftrightarrow$$ ...


1

The statement is indeed true. I don't know how one can prove it rigorously. However, here is some further insight: If you have with you a general equation in x, y, or more variables with some constants a, b, ... Suppose there are n constants in the equation. What would you do if you wanted to find the differential equation that this curve satisfies? You ...


2

Just plugin the definition of $T$ and $Q_\sigma$. We have $$ u = \sigma Tu \iff Tu = \frac{u}{\sigma} $$ that is, by definition of $T$ iff $v := \frac{u}\sigma$ is the (unique) solution of $$ a^{ij}(x,u,Du)D_{ij}v + b(x,u,Du) = 0, \quad v|_{\partial \Omega} = \phi $$ Let here $v = \frac u\sigma$, this gives $$ \frac 1\sigma a^{ij}(x,u,Du) D_{ij}u + ...


1

Concerning the first part, it is just a change of variable. If you define $$\sqrt{y(x)\over k-y(x)}=\tan(\phi(x))$$, you then have $$y(x)=\frac{k \tan ^2(\phi (x))}{1+\tan ^2(\phi (x))}$$ $$y'(x)=k\, \phi '(x)\, \sin (2 \phi (x))$$ Replacing in the original equation, you then arrive to $$k\, \phi '(x)\, \sin (2 \phi (x))=\frac{1}{2} \sqrt{k^2\, \sin ^2(2 ...


1

Note that, with the fact that $a \neq b$, $$\frac{\log(b)}{\log(\frac{b}{a})} - \frac{\log(a)}{\log(\frac{b}{a})} = \frac{\log(b) - \log(a)}{\log(b) - \log(a)} = 1$$ This tells that $$\frac{n\pi\log(b)}{\log(\frac{b}{a})} = n\pi +\frac{n\pi\log(a)}{\log(\frac{b}{a})}$$ and that $$\cos(\frac{n\pi\log(b)}{\log(\frac{b}{a})}) = ...


0

$\phi$ depends on $y$, but $y$ depends on $x$, so I think it means that $phi$ depends on $x$ implicitly, and as far as I can tell, there is no relation between $\phi$ and $\theta$; it's just a integration technique.


0

The coefficient matrix $A$ has characteristic polynomial \begin{align} \left|\begin{array}{cc} \lambda+3 & 5 \\ -3 & \lambda-1 \end{array}\right| & = (\lambda+3)(\lambda-1)+15 \\ & = \lambda^{2}+2\lambda+12 \\ & = (\lambda+1)^{2}+11 \\ & = ...



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