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1

If $a \neq 0$ then you can just separate variables, at least for solutions going forward in time. You can also just separate variables backward in time up until the point where $y$ hits zero. Once $y$ hits zero, separation of variables (or the corresponding technique using the chain rule and no explicit differentials) is no longer guaranteed to work. ...


0

$$\frac{dy}{2y^{1/2}} = dx$$ By integrating from two sides : $$y^{1/2} = x + c$$ $$y = (x+c)^2$$ And since $y(0) = 0$, we deduce $0 = (0+c)^2$, hence $c=0$ So , there is only one answer for $y$. The answer is $y = x^2$ For $y(0) = a$ with $a\ge0$ $a = c^2$, hence $c=\pm\sqrt a$ two answers for $c$, so two answers for $y$


1

Yes, it's classified correctly. The independent variable can be any of your choosing as y is not explicitly stated as a function of a certain variable as it is in say $ay''(x) + by'(x) + cy(x)=0$ where $x$ is the independent variable. Here's a link to some videos that could help refresh your memory! You can go one step further and say that it's also ...


0

Notice, $$\frac{dK}{dt}=\frac{R}{V}K_{in}-\frac{R}{V}K$$ $$\implies \frac{dK}{dt}=\frac{R}{V}(K_{in}-K)$$ $$ \frac{dK}{K-(K_{in})}=-\frac{R}{V}dt$$ Since, $K_{in}$, concentration sands flowing in, is constant hence integrating both the sides w.r.t. time $t$, we get $$ \int\frac{dK}{(K-K_{in})}=-\int \frac{R}{V}dt$$ $$\ln(K-K_{in})=-\frac{R}{V}t+c$$ ...


2

Substitution: $K_{1} = K - K_{in} $. Then equation is $$\frac{d K_1}{d t} = -\frac{R}{V}K_1$$ Solution is $ K_1(t) = C * e^{-\frac{R}{V} t}$ where C is constant difined from boundaries i.e. from t=0. Then $$ K = K_{in} + C * e^{-\frac{R}{V} t} $$ In long run $ K = K_{in} $. NB: Solution assumes that $ K_{in} $ is constant and doesn't depend from t.


0

HINT: You can use Peano theorem to prove (local) existence of solution for your ODE: If $D\in \mathbb{R} \times \mathbb{R}$ is an open subset of $\mathbb{R}^2$, and if $f:D \to \mathbb{R}$ is continuous, then the ODE $$ \begin{cases} y'(t) = f\big(t, y(t) \big), \\ y\left(t_0\right) = y_0, & \left(t_0, y_0\right) \in D, \end{cases} $$ has a ...


1

It's not true. The general solution to your differential equation is $$y(t) = \dfrac{-\exp(-e^t)}{\int \exp(-e^t)\cos(t)\; dt}$$ where the denominator is any antiderivative of $\exp(-e^t)\cos(t)\; dt$. In particular, taking an antiderivative that is $0$ at, say, $t=1$, you get a solution that becomes infinite at that value of $t$. Numerically, this ...


1

Your question is very general, but I think it's possible to give recommendations that work in many cases. Although they are quite general too :) There are two methods for solving such problems that immediately come to my mind. The first one is finding trapping region and the second one uses the existence of first integral. I should note that the first ...


1

$$\left(xy\log\frac{x}{y}\right)dx+\left(y^2-x^2\log\frac{x}{y}\right)=0$$ $$\left(xy\log\frac{y}{x}\right)dx=\left(y^2+x^2\log\frac{y}{x}\right)dy$$ $$\left(\frac{y}{x}\log\frac{y}{x}\right)dx=\left(\left(\frac{y}{x}\right)^2+\log\frac{y}{x}\right)dy$$ Let $\frac{y}{x}=u \implies y=ux \implies \frac{dy}{dx}=u+x\frac{du}{dx}$ $$\left(u\log ...


3

In the form it is in right now, I do not think the Laplace transform is of use. However, you can rewrite it in such a way to use the Laplace transform: $$-y\sin x + y'\cos x = \cos^2 x \Longrightarrow \frac{d}{dx}(y\cos x) = \cos^2 x.$$ From here you can use a Laplace transform. You might want to use a double angle identity to simplify $\cos^2 x$. That ...


0

Your function is $f(t)=t$ but not $f(t)=1$. You need to use the formula $$\mathcal{L}\{u_c(t) f(t-c)\} = e^{-cs}*F(s)$$ 3 times with $c=1,2, 3$ to get the answer. The answer derived by Leucippus looks correct.


0

It might be helpful to consider it as three separate parts: $$h(t) = \left\{\begin{array}{l} 0,\, 0\leq t<\pi \\ 1,\, \pi\leq t<2\pi\\ 0, t\geq2\pi \end{array}\right.$$ The most systematic way I find to turn these into heaviside functions is to start with the top-most, then add heaviside times the function on the next step minus that of the ...


0

This is the chain rule. You're forming the derivative of $z$ with respect to $y$. But a prime is usually used to denote the derivative with respect to $x$. (Indeed the differential equation wouldn't make sense if the prime were intended to denote differentiation with respect to $y$, since $y'$ would then just be $1$.) So $z'$ is the derivative of $z$ with ...


1

Only to make more obvious what was already obvious, as said in several comments : $$z=y'-\sin(x)$$ $$z'=y''-\cos(x)x'$$ $$z'=y-z=y''-\cos(x)x'=y-(y'-\sin(x))$$ $$y''+y'-y-\cos(x)x'-\sin(x)=0$$ $$x'''+x''-x'-\cos(x)x'-\sin(x)=0$$ Numerical computation of $x(t)$ , $y(t)$ , $z(t)$ requires to state a third condition, for example $x''(0)$ , or $z(0)$, any ...


0

If the system is at rest and in equilibrium, then $\ddot\theta=0=\dot\theta$ so $\theta=\frac{\tau}{k}$


0

By definition of the equilibrium point, its time-derivatives must vanish (the equilibrium point doesn't change with time), so $\ddot{\theta}_\mbox{eq} = \dot{\theta}_\mbox{eq} = 0$ and the differential equation then tells you that $\theta_\mbox{eq} = \tau/\kappa$. Added after more comment exchanges: Since you have $(t,\theta)$ points, you can estimate ...


2

Assuming that it should really be $$\vec{\alpha} = \frac{m(x)}{x^2}\,\vec{x}$$ (see my comments below the question), where $x = |\vec{x}| = \sqrt{x_1^2 + x_2^2}$, then $$\alpha_1 = \frac{m(x)}{x^2}\,x_1$$ and $$\alpha_2 = \frac{m(x)}{x^2}\,x_2$$ So, $$ \frac{\partial\alpha_1}{\partial x_1} = \frac{\partial}{\partial x_1}\big(\frac{m(x)}{x^2}\,x_1\big) = ...


6

HINT Separation of variables yields $$ \int \frac{f'(x)dx}{f(x)} = \int (\cos x + \tan x) dx $$ and LHS integrates to $\ln f(x) + C$.


0

The general solution of the following linear system $\dot{x(t)}=Ax(t)+Bu(t);x(t_0)=x_0$ is given by $x(t)=e^{A(t-t_0)}x_0+\int_{t_0}^t \! e^{A(t-\tau)}Bu(\tau) \, \mathrm{d}\tau.$ This can be proved via Laplace transform method. For your system the input $u(.)\in\mathbb{R}$ can be represented by the step function.


0

Let us temporarily assume that the solution can be extended over $R^+$. Next, apply the Laplace transform to the DE; $(s^2U(s)-su(0)-u_x(0))-m^2U(s)=e^{-sx_0}$ Where, $U(s):= \int_0^\infty \! e^{-sx}u(x) \, \mathrm{d}x.$ $\Rightarrow$ $(s^2U(s)-u_x(0))-m^2U(s)=e^{-sx_0}$$\Rightarrow$$U(s)=\frac{u_x(0)+e^{-sx_0}}{s^2-m^2}$ Applying the inverse ...


1

From the result you obtained, it is quite clear that there is a first change of variable $y=\frac 1z$ which, after simplification gives $$2z'-x z-2=0$$ Integrating $2z'-xz=0$ gives $$z=C e^{\frac{x^2}{4}}$$ and then $$C'=e^{-\frac{x^2}{4}}$$ that is to say $$C=\sqrt{\pi }\, \text{erf}\left(\frac{x}{2}\right)+K$$ and then the result.


4

What you're looking for is an integrating factor. If you multiply the whole equation by $e^{-x^2/4}$ observe that you can make the simplification $$ e^{-x^2/4}y' + \frac{1}{2}xe^{x^2/4} y + e^{x^2/4}y^2 \;\; \Longrightarrow\;\; e^{-x^2/4}\frac{y'}{y^2} + \frac{1}{2}x e^{-x^2/4} \frac{1}{y} + e^{-x^2/4} \;\; =\;\; 0. $$ Observe now that we can simplify ...


0

The equation $$\frac{dy}{dt} = \frac{t}{y-2}$$ can be seen in the form \begin{align} (y-2) \, dy = t \, dt \end{align} and upon integration of both sides becomes \begin{align} \frac{y^{2}}{2} - 2 y = \frac{t^{2}}{2} + \frac{c_{0}}{2} \end{align} or $y^{2} - 4 y -(t^{2} + c_{0}) = 0$. Solving this quadratic equation the value of $y$ is of the form ...


1

Complete the square in $y$:$$y^2-4y-t^2=C\\y^2-4y+4=t^2+C+4\\(y-2)^2=t^2+C\\y=2\pm\sqrt{t^2+C}$$ Note we can absorb the $4$ into our arbitrary constant $C$ which parameterizes our family of solutions. From the original equation, notice: $$(y-2)\frac{dy}{dt}=t$$ Let $z=y-2$ so $dz=dy$ giving $$z\frac{dz}{dt}=t\\2z\frac{dz}{dt}=2t\\\int ...


0

As explained in comments, a non-periodic solution whose limit set is a periodic orbit is something spiraling toward a periodic orbit (either from the inside or from outside). An example of such a system is $$\begin{split} x' &= x\cos(x^2+y^2) - y \\ y' &= y\cos(x^2+y^2)+x\end{split}$$ The plot below (courtesy of Wolfram Alpha) shows two periodic ...


0

Using the notation $u_{c}(t) = H(t-c)$, where $H(t)$ is the Heaviside step function, then $$g(t)= (t-1) u_1(t) - 2(t-2) u_2(t) + (t-3) u_3(t)$$ is evaluated as follows. Let $g(s) = \mathcal{L}\{g(t)\}$ such that: \begin{align} g(s) &= \int_{0}^{\infty} e^{-s t} \left[ (t-1) \, H(t-1) - 2 \, (t-2) \, H(t-2) + (t-3) \, H(t-3) \right] \, dt \\ &= ...


0

HINT: $$\mathscr{L}(tu(t))(s)=\int_0^{\infty}te^{-st}dt=\frac1{s^2}$$ Thus, $$\mathscr{L}((t-c)u(t-c))(s)=\int_c^{\infty}(t-c)e^{-st}dt=e^{-sc}\frac1{s^2}$$ NOTE: We derive the result for the Laplace Transform of $t$. To that end, we write $$\mathscr{L}(t)(s)=\int_0^{\infty}te^{-st}dt$$ Now, integrate by parts with $u=t$ and $v=-e^{-st}/s$. Then, ...


0

You may use naive methods. Form of this equation tell me: bro, use power. Therefore I use $y=Ax^n$ and $$ An(n-1)x^{n-2}=x\cdot Ax^n\cdot Anx^{n-1}. $$ First of all, $n-2=1 + n + (n-1)\Longrightarrow n=-2$. Then $An(n-1)=nA^2$, and $A = 0$ or $A=-3$. So, $y=0$ and $y=-3/x^2$ are solutions. But as you can see, $y=\mathrm{const}$ is a solution, not only zero.


1

A lot of this is shorthand that is justified by defining "differentials" in a certain way. A truly rigorous treatment uses methods of differential geometry, for which a good book might be Weintraub's "Differential Forms". The equivalence of $$\tag 1\frac{dy}{dx}=\frac{M(x,y)}{N(x,y)}$$ and $$\tag 2-M(x,y)dx+N(x,y)dy=0$$ is by no means obvious. $(1)$ ...


2

For your Last expression see: Note that: $$\mathrm {d} f(x,y)=\frac{\partial f}{\partial x}.\mathrm {dx}+\frac{\partial f}{\partial y}.\mathrm {dy}$$ $$\mathrm {d} \sqrt{x^2+y^2}=\mathrm {d}(x^2+y^2)^{\frac{1}{2}}\\ =\frac{1}{2}(x^2+y^2)^{\frac{-1}{2}}\cdot\mathrm {d}(x^2+y^2)\\ =\frac{1}{2\sqrt{(x^2+y^2)}}\cdot d(x^2+y^2)$$ So, $$\pm\int ...


0

if this is your equation $$x^2ydy-(xdy-ydx)=0$$ i would write $$x^2y\frac{dy}{dx}-x\frac{dy}{dx}+y=0$$ or $$x^2y(x)y'(x)-xy'(x)+y(x)=0$$


1

"dy/dx" is NOT a fraction but is defined as the limit of a fraction, the "difference quotient". A result of that is that the derivative can often be treated like a fraction. For example, if y is a function of u and u is itself a function of x, then y can be considered a function of x and, by the "chain rule", dy/dx= (dy/du)(du/dx). We cannot prove that by ...


0

for basics of trigonometry you can go for S.L. lony bok, and the topics what you have mentioned i will suggest two books 1) advanced engineering mathematics by erwin kreyszig 2) advanced engineering mathematics by ramana the first one is for deep understanding of the topics and second one is for practice. also if you are still looking for further basics in ...


2

$\cos y=0$ means $y=\pi/2+k\pi$, but these are not solutions because $\tan y$ on the right hand side is not defined for $y=\pi/2+k\pi$: as you know, $\tan y=\sin y/\cos y$, so $\tan y$ is not defined when $\cos y=0$.


3

The general Sturm Liouville eigenvalue problem is $$ -\frac{d}{dx}\left(p\frac{df}{dx}\right)+qf = \lambda wf. $$ Here it is assumed that $w > 0$, $p > 0$. In this context, one must use the weighted $L_w^2$ space with inner product $$ (f,g)_{w}=\int_{I} f \overline{g}w dx. $$ The actual operator which is symmetric in this case ...


0

Hint: $$\frac{dy}{dx}=\frac{\cos^2y\tan y}{1+x^2}$$ $$\frac{dy}{\cos^2 y\tan y}=\frac{dx}{1+x^2}$$ Integrating both the sides we get $$\int \frac{dy}{\cos^2 y\tan y}=\int \frac{dx}{1+x^2}$$ $$\int \frac{\sec^2y dy}{\tan y}=\int \frac{dx}{1+x^2}$$let $\tan y=t\implies \sec^2ydy=dt$, we get $$\ln (\tan y)=\tan^{-1}(x)+C$$ Edit: Notice, in right side of above ...


3

First of all your factorization is wrong: $k_1=1, k_2=2$,so general solution will be, $$y_g = C_1e^x + C_2e^{2x}$$ Next, $$y_p = Ax^3 + Bx^2 +Cx + D, \\ y'_p = 3Ax^2 + 2Bx + C, \\ y''_p = 6Ax + 2B$$and using all this you will get, $$\begin{cases} 2A = 2, \\ -9A+2B = 0, \\ 6A-6B+2C = 0, \\ 2B-3C+2D = -30 \end{cases} \implies \begin{cases} A = 1, \\ B ...


4

The equations in the question can be written as {D[p[x, t], t] + a D[p[x, t], x] - d/2 D[p[x, t], {x, 2}] == 0, p[x, 0] == 0, a p[0, t] - d/2 (D[p[x, t], x] /. x -> 0) - Sin[t] == 0, D[p[xm, t], t] + a (D[p[x, t], x] /. x -> xm) == 0}; Unfortunately, NDSolve[%, p, {t, 0, 1}, {x, 0, xm}] returns unevaluated with the error message, ...


2

The equations you should have are: $2A=2 \\ -9A+2B=0 \\ 6A-6B+2C=0 \\ 2B-3C+2D=-30 \\ \text{ Solving this system should give you your solution. }$


1

I'm not sure if this really counts as an answer, but I'd like to see if this work according to @Michael's sketch is along the right track. As he mentions, as step 1 we homogenize units by setting $w(x)=x^2y(x)$, and obtain a differential equation for $w$: $$ x^2w'' = -6w-2w^2+4xw'+xww' .$$ Then (step 2) we change coordinates to eliminate the explicit ...


2

$\frac{dP}{dt}$ measures the rate of increase of $P$ in currency units per year At any instant the total in the account is increasing at a rate of $rP$ per year and decreasing at a rate of $w$ per year.


3

$r$ is a rate of interest, meaning a percentage per unit time. So say $r$ is 5% per year, the instantatneous rate of growth of the account is 5% of the amount in the account per year: $rP$. Similarly you have to interpret $w$ as the withdrawal rate per unit time, for example, $1,000 per year. You are right that the equations are meaningless if yo take ...


0

$w$ is the (continuous, constant) rate of withdrawal, and $rP$ is the (continuous, proportional) rate of interest accrual.


0

The proof that you are looking for can be found in the page 258 of the Evans book:


0

Write $u=G+y$, where $G=\frac{1}{4\pi |x-x_0|}$ solves $-u_{xx}=\delta(x-x_0)$ in the whole $\mathbb R^1$. Then plugging this in your equation gives $-y_{xx}+m^2G+m^2y=0\quad (*)$ with new boundary conditions: $y(0)=-G(0)$ and $y(L)=-G(L)$. Now you have to solve the regularized equation $(*)$ for $y$.


0

Many trig formulas are things you just look up (or memorize). They can also be sometimes be derived from each other. For example, knowing $\sin^2 x + \cos^2 x =1$, dividing by $\cos^2 x$ yields $\tan^2 x + 1 = \sec^2 x.$ A useful exercise would be to take a list of rules (maybe from your textbook) and see which ones can be derived from other ones. It ...


1

This model is well know as "double-well" potential. The closed form solution to $\mathbb{E}(X(t))$ is not known. See Iacus, ch. 1.13.8. Physically, you would expect that the global solution exists and is unique, since the drift term pushes the particle back to area around the origin every time $X(t)^2 > \tfrac{8}{9}$. We can proceed with simulation, ...


0

Yes, you get $dy/y+ 2u/(3u^2+ 1)du= 0$. The integral of $dy/y$ is $ln(|y|)$ and the integral of $2u/(3u^2+ 1)du$, using the substitution $v= 3u^2+ 1$ so that $dv= 6udu$ or $2udu= dv/3$, is the integral of $1/v dv/3$ which is $$(1/3)ln(|v|)= (1/3)ln(3u^2+ 1)= (1/3)ln(3x^2/y^2+ 1)$$ So integrating $dy/y+ 2u/(3u^2+ 1)= 0$ becomes $$ln(|y|)+ (1/3)ln(3x^2/y^2+ ...


2

The variables $y$ and $u$ can be separated from each other: $$ 2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0 $$ $$ 2u\left( du + u \, \frac{dy} y \right) + (u^2+1 ) \, \frac{dy} y = 0 $$ $$ 2u\left( \frac{du} u + \frac{dy} y \right) + \left( u + \frac 1 u \right) \, \frac{dy} y = 0 $$ $$ 2 \left( \frac{du} u + \frac{dy} y \right) + \left( 1 + \frac 1 {u^2} ...


2

$$2xy\mathrm dx+(x^2+y^2)\mathrm dy = 0$$ $$2xy \mathrm dx=-(x^2+y^2)\mathrm dy$$ $$\frac{dx}{dy}=-\frac{(x^2+y^2)}{2xy}$$ Now put, $x=yu \Rightarrow \mathrm dx=u\mathrm dy+y\mathrm du$, i.e. $\frac {dx}{dy}=u+y\frac{du}{dy}$ Using these substitution and after simplification a you will get that $$\frac{2u}{3u^2+1} \mathrm du=-\frac{1}{y}\mathrm dy$$ ...



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