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Here is another answer that is more straightforward. $\mu(t,x)=1,\,\sigma(t,x)=2\sqrt{x}$. For any $\alpha\ge 0$, $\tau_\alpha:=\inf\{t\ge\alpha: B_t=0\}$, by Ito's Lemma, $$X_t(\alpha)=\begin{cases} 0, & t<\tau \\ B_t^2, &t\ge\tau \end{cases}$$ is a solution of the above stochastic differential equation.

0

Assume $h\neq0$ for the key case: Hint: Let $u=x^2+y^2$ , Then $\dfrac{du}{dx}=2x+2y\dfrac{dy}{dx}$ $\therefore\left(\dfrac{x}{2y}\left(\dfrac{du}{dx}-2x\right)-y\right)\dfrac{1}{2}\dfrac{du}{dx}=\dfrac{h^2}{2y}\left(\dfrac{du}{dx}-2x\right)$ $\left(x\left(\dfrac{du}{dx}-2x\right)-2y^2\right)\dfrac{du}{dx}=2h^2\left(\dfrac{du}{dx}-2x\right)$ $\left(x\... 0 The solution of the equation is below :$$\cos (x)\cdot y'+\sin (x)\cdot y=2(\cos (x))^{ 3 }\cdot \sin (x)-1\\ \cos (x)\cdot y'+\sin (x)\cdot y=0\\ \cos (x)\cdot y'=-\sin (x)\cdot y\\ \int { \frac { dy }{ y } } =-\int { \frac { \sin { x } }{ \cos { x } } dx } \\ \ln { y } =\ln { C\cos { x } } \\ y=C\cos { x } \\ y=C\left( x \right) \cos { x } \\ { ... 1 To be clear, you want to find P_n(t) for a birth-death process with constant birth rate q(i,i+1)=\lambda and constant death rate q(i,i-1)=\mu? This particular birth and death process is exactly the M/M/1 queue. As you can see, under "Transient solution", there is a solution for the probability mass function dependent on time for a particular state. 1 Using the substitution z=x y, y'= \dfrac{xz'-z}{x^2}, gives us$$(x+1)z\dfrac{xz'-z}{x^2} - z=1$$or$$-(x+1)z^2 + x(x+1)z'- x^2z=x^2$$Which implies$$z'=-\dfrac{z^2}{x}+\dfrac{xz}{x+1}+\dfrac{x}{x+1}$$Which is a Riccati equation, which tend to be horrible to solve by hand. Wolfram|Alpha also doesn't give a closed form solution. Maybe someone else ... 1$$(x^2-1)\ddot y-2x\dot y +2y=1$$An obvious particular solution is y=\frac{1}{2} Add it to the solution of the homogeneous ODE :$$y=A(x^2+1)+Bx+\frac{1}{2}$$0 Set y=u+\frac{1}{2} we have$$(x^2-1)u''-2xu' +1(1+1)u=0$$This equation is Legendre Differential Equation. 2 I presume you mean (x^2- 1)y''- 2xy'- y= 1. An obvious thing to try, since the right hand side is a constant, is y= constant. With y= C, y'= y''= 0 so the equation becomes -C= 1 or C= -1. y(x)= -1 satisfies this equation. 1 Differentiate both sides of the equation f(\lambda x) = \lambda^{a}f(x) in \lambda to obtain: $$\sum_{i = 1}^{n}x_{i}(\partial_{i}f)(\lambda x) = a\lambda^{a-1}f(x)$$ Setting \lambda = 1 yields \sum_{i = 1}^{n}x_{i}(\partial_{i}f)(x) = af(x), known as Euler's identity. Differentiating the equation displayed above in \... 2 The quantity F_{y'y'}F_{z'z'}-(F_{y'z'})^2 is the determinant of the matrix associated to your system of equations. The authors consider this system as a linear system in the unknown y'' and z'', and the only solution is the trivial one as soon as the determinant is nonzero. 1 Just for the sake of completion to show that both the formula and the book method give the same answer: The formula:$$y=e^{-I}\int Q(x)e^I\,dx+Ce^{-I}\tag{3}$$gives the correct solution provided that I=\int P(x)\,dx. Then \displaystyle\int\frac2xdx=2\ln|x| By insertion of I and Q(x) into (3):$$y=e^{-2\ln|x|}\int \frac{e^x}{x^2}e^{2\ln|x|}... 0 $$\frac{d^2x}{dt^2}+t^2x=0$$ Changing$t$to$-t$doesn't change$x(t)$. Hense$x(t)$is an even function. The general solution can be expressed in terms of particular parabolic cylinder functions. But the variable of those functions are on the complex range, which leads to more complicated interpretation. The ODE is also a generalized form of Bessel ODE. ... 2 Hint: Assume such a trajectory exists, and consider what this would imply for uniqueness of solutions of the system$x'=-F$,$y'=-G$. 2 I know no other way than trial and error. The factor $$xy'(x)-y$$ suggests the substitution $$u(x)=\frac{y(x)}{x}$$ but that leads to a dead end (try it out!). The factor $$y(x)y'(x)+x$$ suggests the substitution $$u(x)=y(x)^2+x^2.$$ This leads to $$u(x)=-\frac{a^2}{4}+\frac{a^2x}{2u'(x)}+\frac{x}{2}u'(x),$$ which is not in the Clairaut form ... 3 Equation$(3)$is wrong, it must be $$y=\frac{1}{\mu(x)}\left[\int \mu(x)Q(x)dx+C\right]\tag{3}$$ where $$\mu(x)=e^{\int\frac2{x}dx}$$ The book is not wrong, notice that $$\frac{dI}{dx}=P\qquad\implies\qquad I=\int Pdx$$ According to the book we have $$I=\int Pdx=\int\frac2xdx=2\ln|x|+c$$ while you are taking$I=e^{\int Pdx}$0 Indeed, the first step you described is correct, so $$x_1(t) = Ae^{-t}$$ and we have $$x_2'(t) + x_2(t) = A^2 e^{-2t}$$ so you will have to propose the solution $$x_2(t) = Be^{-t} + Ce^{-2t}$$ where the first term is from the homogeneous equation and the second - a solution of the particular form the RHS needs. To find$C$, plug this form into the ODE for$...

2

Note that $$x^r=\left(e^{\ln x}\right)^r=e^{r\ln x}$$ Therefore $$\frac{\partial}{\partial r}x^r=(\ln x) e^{r\ln x}=x^r\ln x$$

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$$y'(x)\ln^2(y(x))=y(x)x^2\Longleftrightarrow$$ $$\frac{y'(x)\ln^2(y(x))}{y(x)}=x^2\Longleftrightarrow$$ $$\int\frac{y'(x)\ln^2(y(x))}{y(x)}\space\text{d}x=\int x^2\space\text{d}x\Longleftrightarrow$$ For the LHS, subtitute $u=\ln(y(x))$ and $\text{d}u=\frac{y'(x)}{y(x)}\space\text{d}x$. $$\frac{\ln^3(y(x))}{3}=\frac{x^3}{3}+\text{C}$$ Now, use $y(1)=... 0 Divide$x^2$from both sides and then let$z=\dfrac{y}x$. Note that$y'=z+xz'$. The new equation becomes$(1+z)y'=\sqrt{1-z^2}+z+z^2$.$y'=\sqrt{\dfrac{1-z}{1+z}}+zx\dfrac{\mathrm dz}{\mathrm dx}=\sqrt{\dfrac{1-z}{1+z}}\dfrac{\mathrm dx}{x}=\sqrt{\dfrac{1+z}{1-z}}\ \mathrm dz\ln(x)=\sqrt{\dfrac1{1-z}}\left((z-1)\sqrt{z+1}+2\sqrt{1-z}\sin^{-1}\...

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Divide both side to the $x^{ 2 }$ $$\left( x^{ 2 }+xy \right) y'=x\sqrt { x^{ 2 }-y^{ 2 } } +xy+y^{ 2 }\\ \left( 1+\frac { y }{ x } \right) { y }^{ \prime }=\sqrt { 1-\frac { { y }^{ 2 } }{ { x }^{ 2 } } } +\frac { y }{ x } +\frac { { y }^{ 2 } }{ { x }^{ 2 } } \\ y=zx\\ { y }^{ \prime }={ z }^{ \prime }x+z\\ \left( 1+z \right) \left( { z }^{ \prime }x+... 0 First of all, the C in your first equation is not the C in your second equation. (If the first C was called C_1, the second C would be equal to 3C_1.) You have y(1)=e^2. Plugging x=1 and y=e^2 into your last equation yields e^2=e^\sqrt[3]{1^2+C}, thus, 2=\sqrt[3]{1+C}. Can you do the rest? 0 It may be easier to solve for C at the step$$(\ln y)^3=x^3+C2^3=1^3+C$$. Everything else looks good. 3 Check out Girsanov's paper from 1962 (here or here) for the proof of why his counterexample works; he gives the counterexample:$$ dX_t = \frac{|X_t|^{\alpha}}{1+|X_t|^{\alpha}}dB_t, \quad 0\le \alpha < \frac{1}{2} $$where B_t is Brownian motion/standard Wiener process. 1 Letting y=\frac{u(x)}{xe^{3x}} gives$$ u''=18xe^{3x}. $$Thus$$ u'(x)=\int 18xe^{3x}dx=2(3x-1)e^{3x}+C_1$$and hence$$ u(x)=\int 2(3x-1)e^{3x}+C_1x+C_2=2e^{3x}(x-\frac23)+C_1x+C_2.$$Therefore$$ y=\frac{1}{xe^{3x}}(2e^{3x}(x-\frac23)+C_1x+C_2). $$1 Justification of the formula: After centering (translation to let the linear terms vanish), the equation becomes$$Ax^2+Bxy+Cy^2+F'=0.$$Then you apply a rotation of angle \theta to let the mixed term Bxy vanish from the quadratic terms,$$A(x\cos\theta-y\sin\theta)^2+B(x\cos\theta-y\sin\theta)(x\sin\theta+y\cos\theta)+C(x\sin\theta+y\cos\theta)^2.$$... 1 Since a particular solution e^{-3x} of the associated homogeneous ODE (i.e.: with term on the right =0 instead of 18x), has already been found the problem is much simplified. This draw us to change of function : \quad y(x)=f(x)e^{-3x} y'=(f'-3f)e^{-3x} y''=(f''-6f'+9f)e^{-3x} x y'' +2(3x+1)y' +3y(3x+2)=\left( x(f''-6f'+9f)+2(3x+1)(f'-3f)+3(... 2 Your original idea is correct. If you substitute u=xy you get$$\frac{du}{dx}=x\frac{dy}{dx}+y$$and$$\frac{d^2u}{dx^2}=x\frac{d^2y}{dx^2}+2\frac{dy}{dx}$$The original differential equation then becomes$$\frac{d^2u}{dx^2}+6\frac{du}{dx}+9u=18x$$Can you finish this now? 1 Use Laplace transform:$$xy''(x)+2y'(x)(3x+1)+3y(x)(3x+2)=18x\Longleftrightarrowxy''(x)+2y'(x)+6xy'(x)+6y(x)+9xy(x)=18x\Longleftrightarrow\mathcal{L}_x\left[xy''(x)+2y'(x)+6xy'(x)+6y(x)+9xy(x)\right]_{(s)}=\mathcal{L}_x\left[18x\right]_{(s)}\Longleftrightarrow$$Now use:$$\mathcal{L}_x\left[y(x)\right]_{(s)}=\text{Y}(s)\mathcal{L}_x\...

1

I see no error here. $$(D-x)(D+x)y_n = y_n^{''}-x^2y_n + y_n = -(2n+1)y_n + y_n = -2ny_n$$ and $$(D+x)(D-x)y_n = y_n^{''}-x^2y_n - y_n = -(2n+1)y_n - y_n = -2ny_n-2y_n = -2(n+1)y_n$$ I'm assuming here that you see why the result in $(22.2)\$ is correct - the operators don't commute. Given that, the results in $(22.3)\$ and $(22.4)\$ correctly follow.

1

Integrating Factor This equation is equivalent to $$(u+v)\,\mathrm{d}u-(u-v)\,\mathrm{d}v=0\tag{1}$$ The integrating factor should be a function $g$ so that when multiplied by $(1)$ gives an exact differential. This is when $$\frac{\partial}{\partial v}[(u+v)g]=\frac{\partial}{\partial u}[-(u-v)g]\tag{2}$$ $(2)$ implies (u-v,u+v)\cdot\nabla\log(g)=-2\... 1 (D - x)(D + x)y = (D - x)(Dy + xy) = (D - x)Dy + (D - x)xy = D^2y - xDy + Dxy - x^2y = D^2y - xDy + (y + xDy) - x^2y = D^2y + y - x^2y Then (D - x)(D + x)y_n = D^2y_n + y_n - x^2y_n = -(2n + 1)y_n + y_n = -2ny_n 1 There're two principal axes in general, so \begin{align*} \theta &=\frac{1}{2} \tan^{-1} \frac{B}{A-C}+\frac{n\pi}{2} \\ &= \tan^{-1} \left( \frac{C-A}{B} \color{red}{\pm} \frac{\sqrt{(A-C)^{2}+B^{2}}}{B} \: \right) \\ \end{align*} The centre is given by(h,k)= \left( \frac{2CD-BE}{B^2-4AC}, \frac{2AE-BD}{B^2-4AC} \right)$$... 1$$u(du-dv)+v(du+dv)=0$$Solving with integrating factor isn't the simplest method in the present case. But it is the required method. So, we have to do it.$$(u+v)du+(-u+v)dv=0$$We look for an integrating factor f(u,v) so that we get the exact differential of a function F(u,v) :$$f(u,v)\left((u+v)du+(-u+v)dv)\right)=dF(u,v)\frac{\partial F}{\...

1

The general solution to $$\frac {dC}{dt}=-0.3C$$ is $$C_t=C_0e^{-0.3t}$$ which is the same as $$t=-\frac 1{0.3} \ln \left(\frac{C_t}{C_0}\right)=\frac 1{0.3}\ln\left(\frac{C_0}{C_t}\right)$$ Apply this for each cup of coffee consumed. First cup: $C_0=20, C_t=12 \Rightarrow t=1.7028$. Second cup: $C_0=20+12=32, C_t=12 \Rightarrow t=3.2694$. Third cup: $... 1 Notice that just because the writer begins another cup, it doesn't mean that there is no caffeine in him. In fact, when he begins cup #2, the writer will have$12 + 20 = 32$milligrams of caffeine in his system. As you can see, it will take longer after he drinks the second cup before he becomes drowsy again. To be honest, you can just do the calculations ... 1 Would it be something like find the time it takes for C to go from 20->12 in the first bout t1 and then 32 to 12 t2 using (0,32) as the initial condition? Then it would be something like t1+n*t2 = 10hr. Solve for n and round down (i think) to find how many cups and add one for t1. This is because it will always go in the pattern 20->12, then 32->12 until end.... 9 No, there is not. Note that if$f$is strictly increasing, then$f'(x)\ge 0$for all real$x$. Suppose$f'(x) = f(f(x))$for all real$x$. Note that if$f(0)<0$, then by strict monotonicity$f(f(0))< f(0) < 0$, and hence$f'(0) < 0$, contradiction. So$f(0)\ge 0$, and in particular$f(x)>0$if$x>0$. As such, choose$c>0\$, and let $$g(... 0 I can't find any real function, but this function is satisfying your equation$$f(x)=cx^{\frac{1+\sqrt{3}i}{2}}$$where c=e^{\frac{\sqrt{3}\pi}{6}}[\frac{\sqrt{3}+i}{2}]. I got this function by equating the degrees on both sides but if you find any real function satisfying this equation, please do let me know. Hope this helps ! 0 Hint:$$\int\dfrac{u'(x)}{u(x)}\,\mathrm{d}x=\begin{cases}\ln\big(u(x)\big)+\mathrm C_1, & u(x)\gt0 \\\ln\big(-u(x)\big)+\mathrm C_2, & u(x)\lt0\end{cases}$$2 Note that the characteristic function of the sum of the random variables being described is:$$(\frac{1}{2}e^{-it}+\frac{1}{2}e^{it})^k=\cosh^k(it)=\cos^k(t)$$since it is the sum of k independent random variables, each of which has the characteristic function \frac{1}{2}e^{-it}+\frac{1}{2}e^{it} (use the convolution formula). Then he just uses one of ... 0 As pointed out by @Ian, the coefficient of the derivative should be 1. Moreover, integrating factor is  e^{\int (coefficient\space of \space x) dt}  which is \frac {1}{t} in this case.\frac{1}{t^2}  must be a typo. 6 You want a coefficient of 1 on the derivative. So your ODE should instead look like \frac{dx}{dt}-\frac{1}{t}x=-t-4/t. Then your integrating factor would be e^{-\ln(t)}=1/t; that converts the equation into \frac{d}{dt}(x/t)=-1-4/t^2. I don't really understand the integrating factor of 1/t^2 here. 1 Although it sounds like a question, for calculation did you use atan2 function or atan function? Quadrant placement is also important. 1 The Lipstcitz condition is not necessary for uniqueness of solutions. There are other sufficient conditions, like the Osgood criterion, or monotony conditions like$$ (f(t,x)-f(t,y)\cdot(x-y)\le0. $$As far as I know, there are not known necessary and sufficient conditions. 0 Assume a\neq0 for the key case: \dfrac{dx}{dt}=\dfrac{at-\cos x}{at^2\tan x+t} (at-\cos x)\dfrac{dt}{dx}=at^2\tan x+t This belongs to an Abel equation of the second kind. Let u=at-\cos x , Then t=\dfrac{u+\cos x}{a} \dfrac{dt}{dx}=\dfrac{1}{a}\dfrac{du}{dx}-\dfrac{\sin x}{a} \therefore\dfrac{u}{a}\dfrac{du}{dx}-\dfrac{u\sin x}{a}=a\dfrac{(... 0 y''-(x+1-\lambda)y=0. You need x+1-\lambda <0 which gives \lambda>x+1 i.e. \lambda>2 for x\in(0,1). 3$$ \dot{y} = \frac{3x^2}{y-x^2+1}$$u=y-x^2+1\quad\to\quad \dot{u}=\dot{y}-2x =\frac{3x^2}{u}-2x$$u\dot{u}=-2xu+3x^2$$This is an Abel's differential equation of the second kind. In the present case, as far as I know, there is no standard special function available to express the general solution on a closed form. Use numerical calculus to solve it. 0 It's been a while since I've done derivatives, been here's my attempt. We're just trying to solve \frac{d}{dx}[f(y)] = y^{2}(3\frac{dy}{dx} + 1) = 3y^{2}\frac{dy}{dx} + y^{2}. Thus, integrate both sides by x to find f(y):$$ \int 3y^{2}\frac{dy}{dx} + y^{2} dx = \int 3y^{2}\frac{dy}{dx} dx + \int y^{2} dx. $$The first integral is, of course, y^{3}... 0 The formula on the left hand side of your picture should be used when P(x,y) can be easily integrated with respect to x  and the formula to the right hand side should be used when Q(x,y) can be easily integrated with respect to y. -1 Use Laplace transform:$$ \begin{cases} y'_0(t)=\left[r_y(1-u)-d_0\right]y_0(t)\\ y'_1(t)=a_yy_0(t)-d_1y_1(t)\\ y'_2(t)=b_yy_1(t)-d_2y_2(t)\\ y'_3(t)=c_yy_2(t)-d_3y_3(t) \end{cases}\Longleftrightarrow \begin{cases} s\text{Y}_0(s)-y_0(0)=r_y(1-u)\text{Y}_0(s)-d_0\text{Y}_0(s)\\ s\text{Y}_1(s)-y_1(0)=a_y\text{Y}_0(s)-d_1\text{Y}_1(s)\\ s\text{Y}_2(s)-y_2(0)=...

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