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The equation is $\displaystyle \frac{dy(x)}{dx}\cdot{\frac{x^2}{y(x)^2}}+\frac{dy(x)} {dx}\cdot{x(x-y(x))}=1$. Multiplying by $y(x)^2$ we get $$\displaystyle \frac{dy(x)}{dx}\left[x^2+x(x-y(x))y(x)^2\right]=y(x)^2 \rightarrow \frac{dy(x)}{dx}=\frac{y(x)^2}{x^2+x^2y(x)^2-xy(x)^3}$$ Now, $\displaystyle \frac{dy(x)}{dx}=\large \frac{1}{\frac{dx(y)}{dy}}$, so ...

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Hint:- $\left(\dfrac{x^2}{y^2} +x(x-y)\right)=\dfrac{dx}{dy}$ $x=vy\implies \dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$ $\therefore \left(\dfrac{x^2}{y^2} +x(x-y)\right)=\dfrac{dx}{dy} \\\implies v+y\dfrac{dv}{dy}=v^2+v^2y^2-vy^2 \\ \implies y\dfrac{dv}{dy}=v^2+v^2y^2-vy^2-v\\ \implies \dfrac{dv}{dy}=(1+y^2)v(v-1) \\\implies \displaystyle ... 0 You start with fixing$t = t_1$. Let$A = y_1(t_1)$. This is a constant now. Then apply the linearity of the integral: $$A \int_{t_0}^{t_1}y_2(s) ds= \int_{t_0}^{t_1}Ay_2(s) ds$$ In other words: $$y_1(t) \int_{t_0}^{t}y_2(s) ds= \int_{t_0}^{t}y_1(t) y_2(s) ds$$ This is true for every$t$, as we did not use any particular property of$t_1$. 0 Here, there is already a problem with the characteristic equation: $$r^2 - 1 = 0 \iff r = \pm 1$$ I recommend you to take a look at the variation of constants: $$y_P = Ae^x + Be^{-x}$$with$A,B$such as $$A'e^x + B'e^{-x} = 0$$ When you write the derivatives you get: $$y_P' = Ae^x - Be^{-x} \\ y_P'' = Ae^x + A'e^x + Be^{-x} - B'e^{-x}$$ so $$... 0 y_p=C_1e^{x}+C_2e^{-x} and for the special solution of the inhomogeneous equation make the ansatz (Ax+B)e^{x} 2 With steplength h, the implicit Euler method for this equation would say$$y_{n+1}-y_n=hy_{n+1}^2.$$Seeing this as a quadratic equation to solve for y_{n+1}, we get$$y_{n+1}=\frac{1\pm\sqrt{1-4hy_n}}{2h},$$but you need to pick the correct sign. For concistency, you absolutely need to have y_{n+1} converging to y_n if you let h\to0, and that ... 1 You have$$ \dot z = f(t) $$where z(t)=\ln(r(t) and f(t)=δ(ωt)\cos(2θ(t))+ϵ(ωt)\sin(2θ(t)) So obviously$$ z(t)=z(\tau)+\int_\tau^t f(s) ds $$Namely$$ z(t)-z(\tau)=\ln(r(t))-\ln(r(\tau)=\ln(\frac{r(t)}{r(\tau)})=\int_\tau^t f(s) ds $$but r(t)=\sqrt{x_1^2(t)+x_2^2(t)}=||x(t)|| The fact that f(s)=f(\omega(s),\phi(s)) follows from the condition ... 0 Just denote x y = u. Then y = \frac{u}{x} and$$ y' = \frac{u' x - u}{x^2}. $$Substitution of these things to the problem leads to$$ \frac{u' x - u}{x^2} = \frac{u}{x^2} \left(\frac{u + 1}{u - 1}\right), $$i.e.$$ u' x - u = \frac{u (u + 1)}{u - 1}, $$i.e.$$ u' x = \frac{2 u^2}{u - 1}. $$Now you can proceed with the standard separation of ... 1 Note that (by the chain rule)$$ \frac{d}{dt}(-1/x) = \frac 1{x^2} \cdot \frac{dx}{dt} = 1 $$1 \def \d{\mathrm{d}} Note that x' \equiv \frac{\d x}{\d t} Then we have$$\frac{\d x}{\d t}=x^2 \iff \frac{1}{x^2} \d x= \d t \implies \int\frac{1}{x^2}\d x=\int 1 \d t .$$You should be able to integrate both sides. 2 Hint: \dfrac{dx}{dt} = x^2 \to \dfrac{dx}{x^2} = dt 0 Upon closer inspection during dinner it appears that a_i = 0 if i is odd, and$$a_i = \frac{a_{i-2}(n-i+2)(n-i+1)}{i}$$if i is even, where a_0=1. 3 Your problem can be reformulated as follows (upon the change of notation a=x, b=y, c=z, d=t). You are assigning the differential form$$ \omega=f_1(x)dx +f_2(y) dy + f_3(z)dz+f_4(t)dt, $$which is closed, hence exact on \mathbb{R}^4. You want to find a potential function, that is, a function F=F(x, y, z, t) such that dF=\omega. One of them is given ... 1 Hint: Try separation of variables$$ \int\limits_{z_0}^z \frac{d\zeta}{\sqrt{1+\zeta^2}} = \int\limits_{x_0}^x d\xi $$then solve for z. -2 the constant function satisfies your equation, what other function could it be? it is a hyperbolic function 0 Let x=\dfrac{1}{u} , Then x'=-\dfrac{u'}{u^2} \therefore-\dfrac{u'}{u^2}=\dfrac{1}{u^2}-\dfrac{2}{t^2} u'=\dfrac{2u^2}{t^2}-1 Let v=\dfrac{u}{t} , Then u=tv u'=tv'+v \therefore tv'+v=2v^2-1 tv'=2v^2-v-1 t\dfrac{dv}{dt}=(2v+1)(v-1) \dfrac{dv}{(2v+1)(v-1)}=\dfrac{dt}{t} \int\dfrac{dv}{(2v+1)(v-1)}=\int\dfrac{dt}{t} ... 1 The function G\colon \mathbb{R}\to \mathbb{R};\; G(u) = f(-u) - f(u) is Lipschitz continuous with G(0) = 0. So the solutions to the differential equation$$y' = G(y)\tag{$\ast$}$$to a given initial condition are unique by the Picard-Lindelöf theorem. It is easily verified that \psi(t) = 0 is a solution of (\ast) for the initial condition y(t_0) ... 0 You have a typo in your first step, you meant sin. But that is not your confusion. I suspect when trying to calculate C and D you forgot that \cos (0) \neq 0. If you did that, you would start off thinking that D=-1 (where actually D=0), and the calculation of C becomes messier. 0 First of all lets' chnage the variables:$$ X=x+y\\ Y=x-y $$Then your equqtion will be converted into two independent equations:$$ i\alpha \frac {dX}{dq}=(\beta -i g)X\\ i\alpha \frac {dY}{dq}=-2c\cos(q)Y\\ $$The both equations can be easily integrated now. I hope you can do it yourself. 1 I have the following:$$x_1:=r(t) \cos \Phi(t),x_2:=r(t) \sin \Phi(t)$$and the derivatives are:$$\dot{x_1}=\dot r \cos \Phi - r \sin(\Phi)\, \dot \Phi,\dot{x_2}=\dot r \sin \Phi + r \cos(\Phi)\, \dot \Phi.$$with \dot x \equiv \dfrac{d x}{dt}. So you get:$$\dot r \cos \Phi - r \sin(\Phi)\, \dot \Phi =a r\cos \Phi+br\sin\Phi,\dot r ... 0$\textbf{hint}$$$f''+\frac{1}{t}f' = \dfrac{tf'' +f'}{t} =\dfrac{1}{t}\dfrac{d}{dt}\left( tf' \right) = 0$$ Can you take it from here? 0 If$x_0=0$, then$x(t)=0$is a solution. Otherwise, I do not see any condition you might impose to guarantee existence. Consider for instance $$f(x,t)=\begin{cases} \phantom{-}x^2 & \text{if x\in\mathbb{Q},}\\ -x^2 & \text{if x\not\in\mathbb{Q}.} \end{cases}$$ Then$|f(x,t)|\le|x|^2$, with$b(t)=1$. 0 at first you must solve the equation$y''-4y'+3y=0$with the ansatz$y=e^{\lambda x}$for the inhomogeneous equation set$y=Ax+B$0 The general method is the following (it is described on wiki): solve the homogeneous equation: $$y'' -4y' + 3y= 0 \implies y = Ae^x + B e^{3x}$$ Now change the unknown function for: $$y(x)= A(x)e^x + B(x)e^{3x}$$ such as $$A'(x)e^x + B'(x)e^{3x} = 0$$and you should find a solution. indeed, when$y$has the previous form: $$y' = Ae^x + 3Be^{3x} ... 0 According to Maple, the first four terms of the series solution of the differential equation around x=0 with initial condition y(0)=y_0, y'(0) = y_1, y''(0)=y_2, are$$y \left( x \right) =y_{{0}}+y_{{1}}x+{\frac {y_{{2}}}{2}}{x}^{2}-{ \frac {{{\rm e}^{ \left( 2+c \right) \ln \left( y_{{2}} \right) }}}{6 }{\frac {1}{\sqrt {y_{{0}}}}}}{x}^{3}+{\frac ... 0 Answer to question (1): When$x\to 0, e^{-x}\to 1$, the leading terms in the equation are: $$y^{1/2}y'''+(y'')^{2+c}=0$$ When$x\to\infty$,$\frac{x}{x+1}\to 1$, so the leading term in the equation are: $$y^{1/2}y'''-yy'=x$$ 2 $$(f(x)+\int_0^1f(x)dx)'=f(x)+\int_0^1f(x)dx$$ $$\Rightarrow f(x)+\int_0^1f(x)dx=e^x[1+\int_0^1f(x)dx]$$ $$\Rightarrow2\int_0^1f(x)dx=(e-1)[1+\int_0^1f(x)dx]$$ $$\Rightarrow\int_0^1f(x)dx=\frac{e-1}{3-e}$$ 3 You get $$f''=f',$$ and thus$f'(x)=c_1e^x$, and hence$f(x)=c_1e^x+c_2$. Next $$f'=f+\int_0^1 f(x)\,dx$$ implies that $$c_1e^x=c_1e^x+c_2+c_1(e-1),$$ and hence$c_2=-c_1(e-1)$, which means that the general expression of$f$is $$f(x)=c_1\big(\mathrm{e}^x-\mathrm{e}+1\big).$$ Incorporating now the initial condition$f(0)=1$, we obtain that ... 4 Let$A=\int_{0}^{1}f(x)dx$. $$f'(x)=f(x)+A$$ The solution is: $$f(x)=-A+c e^x$$ So $$\int_0^1 f(x)dx=-A + (e-1) c=A \implies c=\frac{2A}{e-1}$$ 1 Here is a result, though I think it's not complete. As in my previous solution, I'll consider only the$z$-dependence of each function. Let the differential operator in this problem and the linked question be written as$\mathcal{L}=1-2\lambda D_z$. Then the results of the linked question may be summarized as: If$\mathcal{L}W=G$, then$\displaystyle ...

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If you solve the equation near $t\to\infty$, $$x'(t)+x(t)=\alpha$$ You will obtain: $$x(t)=\alpha+ce^{-t}$$ $$x'(t)=-ce^{-t}$$ So $$\lim_{t\to \infty}x'(t)=0$$ $$\lim_{t\to \infty}x(t)=\alpha$$

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So you're solving for $v_o$, right? Okay. Break the initial vector up into its component vectors $v_{oy}$ and $v_{ox}$. The $x$-component of a vector $V$ at an angle $\theta$ degrees above the horizontal is always $V \cos\theta$, and the $y$-component is always $V \sin\theta$. So your $x$- and $y$- components will be $v_{ox}=v_o\cos\theta$ and ...

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Let $$(\mathcal{L}f)(s)=\int^{\infty}_{0}f(t)e^{-st}\,dt$$ Denote the Laplace operator applied to a real valued function $f$ which satisfies some regularity conditions such that the integral above exists. Note that ...

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$$\underbrace{s^2-10s+29 = (s-5)^2+2^2}_{\text{completing the square}} = t^2+2^2$$ So $$\frac{6s+9}{s^2-10s+29} = \frac{6(s-5)+39}{(s-5)^2+2^2} = \frac{6t+39}{t^2+2^2} = 6\frac{t}{t^2+2^2} + \frac{39}{2}\cdot\frac{2}{t^2+2^2}$$ Now look at each term separately.

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We have $s^2-10s+29=(s-5)^2+2^2$, which implies that $$F(s)=\frac{6s+9}{s^2-10s+29}=\frac{6(s-5)+39}{(s-5)^2+2^2}.$$ Therefore, $$\mathcal{L}^{-1}(F(s))=6\mathcal{L}^{-1}\left(\frac{s-5}{(s-5)^2+2^2}\right) +\frac{39}{2}\mathcal{L}^{-1}\left(\frac{2}{(s-5)^2+2^2}\right)\\ =6e^{5t}\cos(5t)+\frac{39}{2}e^{5t}\sin(2t)$$ by here.

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$$F(s) = \dfrac{6s+9}{s^2-10s+29}=\dfrac{6s-30+39}{(s-5)^2+4}=\dfrac{6s-30}{(s-5)^2+4}+\dfrac{39}{(s-5)^2+4}=6\dfrac{s-5}{(s-5)^2+4}+\dfrac{39}{2}\dfrac{2}{(s-5)^2+4}$$ Then $$\mathcal{L}^{-1}\{F(s)\} =6e^{5t}\cos{2t}+\frac{39}{2}e^{5t}\sin{2t}$$

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Assume that we are interested in stability of the trivial solution $x=0$. We have that there is function $V(x)$ such that $V(x)>0,\,x\in U\backslash\{0\}$, and $\dot V(x)\leq 0,\,x\in U$. Take a ball $B_\epsilon\colon |x|\leq \epsilon\subset U$. Since $V$ is continuous and positive, it reaches a minimum on $\partial B_\epsilon$, call it $k$. Now take ...

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Take for example: $$\left(\frac{y}{p(x)}\right)'=\frac{y}{p(x)}q'(x)$$ or equivalently $$p(x)y'-p'(x)y=p(x)q'(x)y,$$ or $$y'-\left(\frac{p'(x)}{p(x)}-q'(x)\right)y=0.$$

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To permute the rows you just swap them. The matrix with permuted rows is $\left(\matrix{3&-9&-6\\-4&-1&2}\right)$ Now you continue the next step with this new matrix.

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I have redone the calculations and solved the problem. I'll leave my solution here, in case it helps someone. This time, my mistake was not writing the equation as: $$y'' - \frac{2x}{x^2 - 1}y' + \frac{2}{x^2 - 1}y = (x^2 - 1)$$ before applying the method.Really, $y_1 = x$ and $y_2 = x^2+1$ are solutions for the homogeneous equation. Then, we go for: ...

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As a quick method of solving this problem quickly, note that multiplying both sides by $e^{-x}$ yields $$e^{-x}y''-2e^{-x}y'+e^{-x}y=(e^{-x}y)''=1$$ To see this you could note that $$y''-2y'+y=(y'-y)'-(y'-y)$$ for which the appropriate integrating factor would be $e^{-x}$. And after integrating, $y'-y$ has the same integrating factor. Or it's easy to ...

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First, we may note that the $x,y$ dependence is irrelevant, along with the shift in the argument of $g(x,y,z)$. To that end I'll introduce $W(z)=w(x,y,z)$ and $G(z)=g(x,y,z+h)\to g(z)$. Then we just have a simplified ODE given as $(1-\lambda D_z)W(z)=G(z)$. This can indeed be handled by an integrating factor by observing ...

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Let $u(t) = x(t) - t$. Then the equation becomes $$\left\{ \begin{array}{c} \frac{du}{dt} = 1 - \frac{u}{1+2t^2 + 2tu + u^2}\\ u(1) = 0 \end{array} \right.$$ Then since the denominator $1+2t^2 + 2tu + u^2$ is always positive (and in fact greater than 1),$$\lim_{t \rightarrow \infty} u(t) = 0 \ \Rightarrow \lim_{t \rightarrow \infty} \frac{du}{dt} = 1$$ ...

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You're off to a good start. The fact that you ended up with the equation $0=e^x$ isn't due to any calculation error on your part, but rather is because $y(x)=Axe^x+Be^x$ is a solution to the homogeneous system, since then $$y'=Axe^x+Ae^x+Be^x=y+Ae^x$$ and $$y''=y'+Ae^x=y+2Ae^x,$$ so that $$y''-2y'+y=y+2Ae^x-2(y+Ae^x)+y=0.$$ Your general idea is a good one, ...

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Hint: if $\lim t-x(t)=0$, then there exist "arbitrary large values" such that $x'(t)\geq 1/2$. However for sufficiently large $t$, if $t-x(t)\leq 1$ (which forces $x(t)\geq t/2$), then $x'(t)\leq 1/(1+t^2+t^2/4)\ll 1/2$...

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You made an error when you differentiated implicitly. You did not apply the Product Rule $(fg)' = f'g + fg'$ to the term $xy$. Keeping in mind that $y$ is a function of $x$, you should obtain $$(xy)' = 1y + xy' = y + xy'$$ Therefore, when you differentiate $$x^2 + xy + y^2 = 3$$ implicitly with respect to $x$, you should obtain $$2x + y + xy' + 2yy' ... 1 As answered by @MPW: Given$$f(x + \Delta x)=f(x) + a\Delta xf(x)-10b\Delta x$$Which, upon rearrangement yields$$f(x + \Delta x) - f(x) = a\Delta xf(x) - 10b \Delta x$$Dividing through by \Delta x, gives$$\frac{f(x+\Delta x)-f(x)}{\Delta x} =\frac{a\Delta xf(x)-10b\Delta x}{\Delta x}$$Now taking the limit as \Delta x \to 0 yields: ... 1 Hint: Subtract f(x) from both sides and then divide both sides by \Delta x. When you take the limit as \Delta x\to 0, you should recognize what the left side becomes. 0 The first-order system :$$y'=-\frac{x}{c}x'=c(y+x-\frac{x^3}{3})$$leads to :$$x''=c(y'+x'-x^2x')x''=c(-\frac{x}{c}+x'-x^2x')x''+ c(x^2-1)x' + x = 0$$This non-linear second order ODE of autonomous kind can be reduced to a first order non-linear ODE. 2$$(tx)'=tx'+x=tx^2-\frac{2}{t}+x=\frac{(tx-1)(tx+2)}{t}\Rightarrow\frac{(tx)'}{tx-1}-\frac{(tx)'}{tx+2}=\frac{3}{t}\Rightarrow\ln|\frac{tx-1}{tx+2}|=3\ln|t|+\ln|C|\Rightarrow x=\frac{1+2Ct^3}{t(1-Ct^3)}

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