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Idea to try and answer: (might be a bit long for a comment) The system of equations seems to be $$\frac{d y_1}{dt}=f_{12}y_2 \\ \frac{d y_2}{dt}=f_{21}y_1$$ If you differentiate again, you can get second order linear homogeneous differential equations for each function. $$\frac{d^2 y_1}{dt^2}=\frac{1}{f_{12}}\frac{d f_{12}}{dt}\frac{d y_1}{dt}+f_{12}f_{21}... 0 Hint: Let r=t^3 , Then \dfrac{dv}{dt}=\dfrac{dv}{dr}\dfrac{dr}{dt}=3t^2\dfrac{dv}{dr} \dfrac{d^2v}{dt^2}=\dfrac{d}{dt}\left(3t^2\dfrac{dv}{dr}\right)=3t^2\dfrac{d}{dt}\left(\dfrac{dv}{dr}\right)+6t\dfrac{dv}{dr}=3t^2\dfrac{d}{dr}\left(\dfrac{dv}{dr}\right)\dfrac{dr}{dt}+6t\dfrac{dv}{dr}=3t^2\dfrac{d^2v}{dr^2}3t^2+6t\dfrac{dv}{dr}=9t^4\dfrac{d^2v}{dr^2}+... 3$$(3y^2-x)dx + (2y^3-6xy)dy=0$$Another approach (substitute w=y^2):$$(3y^2-x)dx + (y^2-3x)dy^2=0(3w-x)dx + (w-3x)dw=0w'=-\dfrac {(3w-x)}{(w-3x)}$$It's homogeneous. Substitute w=tx. Then it's separable. 0$$\frac{d^2y}{dx^2}=\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}$$You can check the solution just rewrite the DE as:$$\frac{d^2y}{dx^2}=(x)'\frac{1}{y}+x \left (\frac{1}{y}\right )'$$Since (fg)'=f'g+fg' we have:$$y''= \left (\frac xy \right )'y'=\left (\frac xy \right) +C$$0 As a symbolic solution exists, it should be possible to transform the equation and integrate it with relatively elementary means. And indeed, by careful examination one finds that the right side is the derivative of \frac xy, so that a direct integration to$$ y'(x)=\frac{x}{y(x)}+c $$is possible. You could try to find this form of the equation from your ... 0 The fundamental matrix is actually$$\begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} e^{-2t} & 0 \\ 0 & e^{4t} \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix}^{-1}.$$Putting aside the sign error, what you wrote is the product of the first two matrices. The third matrix is needed to change bases to the basis ... 1 We can directly project vector  \vec or  on positive x- axis as$$ \cos \angle rox =(\cos v \cos u) $$It is in effect the same when we take cross product of vectors$$ \vec {ox}= (\cos v \cos u,0,0);\; \vec {or} = (\cos v \cos u,\cos v \sin u,\sin v) \cos w_1=\cos ∠rox =\dfrac{(\cos v \cos u)^2+0+0}{(\cos v \cos u)\cdot (1) }=\cos v \cos u \...

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Hint: I guess that the question is to find the differential equation whose solution is: $$c_1 + y^2 = c_2 x^2$$ Just differentiate twice to find the original DE: $$yy' = c_2 x$$ $$.......$$

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Hint. With $p=(x(t),y(t),z(t)),\ \ v = (v_x(t),v_y(t),v_z(t))$ assuming a dynamic model $$\cases{ \dot v = (0,0,-g)-\frac{F_D}{m}\frac{v}{||v||}\\ \dot p(t) = v(t) }$$ with initial conditions $$p(0) = (0,0,0),\ \ \ v(0) = (v_{x_0},v_{y_0},v_{z_0})$$ Suppose we need to determine the orbit from $p_0=(0,0,0)$ to $p_{goal}=(a,b,H)$ with the restrictions $v_{... 0 $$\frac{d^2x}{dt^2}=2x^3 \implies 2\frac{dx}{dt}\frac{d^2x}{dt^2}=4x^3 \frac{dx}{dx}$$ $$\frac{d}{dt}\left ( \frac{dx}{dt}\right)^2 =2x^3 \frac{dx}{dt}.$$ Integrating w.r.t.$t$, we get $$\int \frac{d}{dt}\left ( \frac{dx}{dt}\right)^2 dt =\int 4 x^3 dx \implies \left(\frac{dx}{dt}\right)^2 = x^4+A \implies \frac{dx}{dt}=\sqrt{x^4+A}$$ $$\implies \int \frac{... 0 The integrand function is$$f(x,x',t)=\left(x'(t) +x^2(t) \right)^2$$Euler-Lagrange equation$$\frac {\partial f}{\partial x}-{\frac {\operatorname {d} }{\operatorname {d} t}}\left({\frac {\partial f}{\partial {x'}}}\right)=0$$leads to the ODE$$4 x(t) \left(x'(t)+x(t)^2\right)-2 \left(x''(t)+2 x(t) x'(t)\right)=0$$expanding and simplifying$$2 x(t)^3-x''(... 1 Answering my own question. We can proceed by putting$t=1$in the known equation $$e^{2tx-t^2}=\sum_{n=1}^\infty\frac{H_n(x).t^n}{n!}$$ we get $$e^{2x-1}=\sum_{n=1}^\infty\frac{H_n(x)}{n!}$$ and $$e^{2x-1}+e^{-2x-1}=\sum_{n=1}^\infty\frac{H_n(x)}{n!}+\sum_{n=1}^\infty\frac{H_n(x)}{n!}(-1)^n$$ $$=2\sum_{n=1}^\infty\frac{H_{2n}(x)}{(2n)!}$$as odd terms will ... 0 Well, this is going to be a bit wordy for a comment, but it's not exactly an answer, either: as stated, this is wrong, already for$n=0$, as others noticed, too. @Ivan Neretin pointed out that there are different definitions of$H_n$: probabilists prefer the ones orthogonal with respect to weight function$e^{-x^2/2}$(because that's up to a factor the ... 1 Your calculus is correct : $$x(y)=y-2\ln|y+1|+c$$ You can check it : $$x'=1-\frac{2}{y+1}$$ $$y'=\frac{1}{x'}=\frac{1}{1-\frac{2}{y+1}}=\frac{y+1}{y-1}\quad\text{is OK.}$$ For information : The calculus of the inverse function$y(x)requires a special function, namely the Lambert-W function : https://mathworld.wolfram.com/LambertW-Function.html $$y(x)=-1-2\:... 1 Using the equation you have derived and plugging c = c_1=c_2, we have$$c\left(\frac{\exp(-t/8)}{4} - \frac{\exp(-t/40)}{20}\right) = 0$$Since c \neq 0, we have$$5\exp(-t/8) = \exp(-t/40)\implies 5 = \exp(t/10)\implies t = 10 \ln 5$$Now you plug that into x(t) and solve for x_{max} = 1000 to get c value 0 Let y=xu , Then y'=xu'+u y''=xu''+u'+u'=xu''+2u' y'''=xu'''+u''+2u''=xu'''+3u'' \therefore x^3(xu'''+3u'')-3x^2(xu''+2u')+x(6-x^2)(xu'+u)-(6-x^2)xu=0 x^4u'''+3x^3u''-3x^3u''-6x^2u'+x^2(6-x^2)u'+x(6-x^2)u-(6-x^2)xu=0 x^4u'''-x^4u'=0 u'''-u'=0 u=C_1e^x+C_2e^{-x}+C_3 \therefore y=C_1xe^x+C_2xe^{-x}+C_3x 1 Let z=x+y , Then y=z-x \dfrac{dy}{dx}=\dfrac{dz}{dx}-1 \therefore\dfrac{dz}{dx}-1=\sin z-e^x \dfrac{dz}{dx}=\sin z+1-e^x Which is similar as Solving ode similar to Adler's equation and follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%... 2 You can still proceed with your method, at least up to the point where you differentiate both sides, and there's no need to extract a recurrence for all the coefficients a_n. Assume a solution of the form$$y=\sum_{n\ge0}a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots$$so that the initial conditions tells us y(0)=1\implies a_0=1. Write the ODE in terms of ... 1 Here's a hint: \cosh(2x) and \sinh(2x) satisfy the ODE y''(x)=4 y(x). Use the recurrence relation H_{m}'(x)=2mH_{m-1}(x) and differentiate the series. If the series satisfies the same ODE, it is a linear multiple of \sinh(2x) or \cosh(2x). It is easy to show that it is indeed proportional to \cosh(2x) using a simple parity argument. Then, all ... 0 The simplest way to find this is to solve the differential equation and then work out it's series expansion. The ODE is easily solved as :$$\frac {dy}{dx}=y^2$$Gives :$$\frac d {dx} \left( \frac 1 y \right) = -1$$and this, with condition y(0)=1 gives a solution :$$y(x)=\frac 1 {1-x} = \sum_{k=0}^{\infty}x^k = 1+x+x^2+x^3+x^4+x^5+\dots2 Use the equation instead... \begin{align*} y' = & y^2 \rightarrow y'(0) = 1^2 = 1 \\ y'' =& (y^2)' = 2 y' y = 2y^3 \Rightarrow y''(0)=2 \\ y''' = & (2 y^3)' = 6 y' y^2 = 6 y^4 \Rightarrow y'''(0) = 6\\ \vdots \end{align*} This way you see that y(x)=y(0) + y'(0) x + \frac 12 y''(0) x^2 + \frac 16 y'''(0)x^3+ \cdots = 1+x+x^2+x^3+ \cdots $$... 1 Your y''=2y is wrong. It should be y''=2yy'. 0 The setup is not quite right. The wording of the problem is a little poor, but ultimately it means that the rate of change of the amount of salt is proportional to the remaining salt and the difference between the concentrations. As such, our equations should read$$\frac{dP}{dt}=k(15-P)\left(3-\frac{P}{10}\right)Comparing to what you had, we have a ... 2 \frac{dP}{dt} = k (15-P) (3- \frac{P}{10}) where k is a constant multiplier. dt = \frac{10}{k(15-P)(30 - P)} dP \displaystyle \int dt = 10 = \frac{1}{k}\int_0^9\frac{10}{k(15-P)(30 - P)} dP = \frac{1}{k} \frac{2}{3} \ln \frac{7}{4} (time taken is 10 mins). So, k = \frac{1}{15} \ln \frac{7}{4} \approx 0.0373 Now for 90\% of salt = 13.5 lb. ... 0 Hint: you can multiply by x to make it Euler Cauchy Differential Equation ,see https://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation 2 You can use the chain rule (derivative of compositions). For instance, \begin{align*} \frac{\partial u}{\partial t} =& \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial t}\\ = &\frac{\sin y}{x} x^{\sin y} \cdot (-3t^2) + \cos y \log x\cdot x^{\sin y} \cdot \frac{s^2}{2\sqrt{t-s}}\\ \... 1 Since you have that m^2=0 \implies m=0 is a double root you have to multiply by \ln |x| in order to find the second solution:y_1=x^0=1\implies y_2= 1 \times \ln |x|y(x)=y_1+y_2 =C_1+C_2 \ln |x|$$2 Using the integrals$$ \int\frac{1}{\sqrt{1-x^2}}= \sin^{-1}x+c_1, \ \ \ \ \int\frac{1}{\sqrt{1-y^2}}= \sin^{-1}y+c_2 $$We get$$ \sin^{-1}x=\sin^{-1}y+c_3$$Thus:$$x=\sin(\sin^{-1}y+c_3)=\sin(\sin^{-1}y)\cos c_3+\cos (\sin^{-1}y)\sin c_3\\=y\cos c_3+\sqrt{1-y^2}\sin c_3 $$Or:$$ (x-y\cos c_3)^2=\sin^2 c_3(1-y^2)  x^2-2xy\cos c_3+y^2\cos^2 c_3=\sin^... 0 Assumek\neq0$for the key cases$x^2y''+xy'+k^2x^2(x^\beta+1)y=a^2yx^2y''+xy'+(k^2x^2(x^\beta+1)-a)y=0$Which is a more generalized version compare to Bessel-like differential equation Some special cases:$\beta=0,-2$: convertible to Bessel ODE$\beta=-1,2$: convertible to degenerate hypergeometric ODE$\beta=-4$: similar to Hunt for exact solutions ... 1 It's $$(xy')'=0$$ or $$xy'=C$$ or $$y=C\ln|x|+C_1.$$ 1 The problem was twofold: 1) I should have used norm2 instead of abs when implementing in FORTRAN 2008. i.e. This line: to_add=(rj-ri)big_gmasses(i)*masses(j)/((abs(rj-ri))**3.0d0) --> to_add=(rj-ri)big_gmasses(i)*masses(j)/((norm2(rj-ri))**3.0d0) My data was messed up, so I tried switching to some one else's, but I forgot to change the units when I did. ... 0 $$\cos x\frac{dy}{dx}+y\sin x=\frac{\cos x}{e^x}$$ $$\left (\frac{y}{\cos x}\right )'=\frac 1{\cos x e^x}$$ There is nothing you can do about the integral on RHS. The integral is hard to calculate. The mehod used to integrate the DE won't chane anything to this problem.Keep the integral in the solution. $$y(x)=C \cos x +\cos x\int \frac {dx}{\cos x e^x}$$ 1 Hint: Multiply both sides of ODE by$\sec^{2}(x)$, so you obtain a exact equation and if considerer the ODE as a linear equation, so you can take$\mu(x)=\sec(x)$as integrate factor. 1 The second case you should be able to solve the DE: $$y''=\tan x$$ Integrate twice. $$y'=c_1-\ln |\cos x|$$ $$y=c_1x+c_2-\int \ln |\cos x|dx$$ You made a mistake maybe you have to consider$\alpha^2 =0,1,-1$. Because$\alpha =-1 \implies \alpha ^2 =1$For the first case: $$y''+y = \tan(x)$$ What you find is correct: $$\lambda^2+1=0 \implies \lambda= \pm ... 0 Hint: x(1-x)(a-b^2x)F'(x)+\dfrac{1}{2}(x(1-x)b)^2F''(x)+k(x-z)^2-rF(x)=0 b^2x^2(x-1)^2F''(x)+2x(x-1)(b^2x-a)F'(x)-2rF(x)=-2k(x-z)^2 F''(x)+\dfrac{2(b^2x-a)}{b^2x(x-1)}F'(x)-\dfrac{2r}{b^2x^2(x-1)^2}F(x)=-\dfrac{2k(x-z)^2}{b^2x^2(x-1)^2} F''(x)+\left(\dfrac{2a}{b^2x}-\dfrac{2(a-b^2)}{b^2(x-1)}\right)F'(x)-\left(\dfrac{2r}{b^2x^2}+\dfrac{4r}{b^2x}-\... 2 Both expressions are equivalent to$$Ce^{(\log x)^2/2}$$or$$C\sqrt{e^{\log^2x}}$$or$$C\sqrt{x^{\log x}}$$or$$C{x^{\log\sqrt x}}$$or$$C{x^{(\log x)/2}}.$$0 You can prove the contrapositive of the statement like this: consider n linearly dependent functions f_i(t),1\le i\le n all n-1 times differentiable in the given domain of t. Then \exists c_1,...,c_n not all 0 such that \forall t, c_1f_1+...+c_nf_n=0\\c_1f_1'+...+c_nf_n'=0\\~~~~\vdots\\c_1f_1^{(n-1)}+...+c_nf^{(n-1)}_n=0 WLOG assume c_1\ne0.... 0 Your equation is y^2\cos x dx + y f(x) dy =0. It is exact if 2y\cos x = yf'(x). Then, f'(x)=2\cos x and f(x) = 2\sin x+c. 0 There are solution x^m of (x^2 f')'+\lambda f=0, where$$ m(m+1)+\lambda=0 \\ m^2+m+\lambda = 0 \\ (m+1/2)^2=1/4-\lambda \\ m=-1/2\pm\sqrt{1/4-\lambda}. $$For convenience, normalize this solution at x=1 by imposing f(1)=0, f'(1)=1, which gives$$ ... 2 The idea to find solutions as exponentials$y=e^u$is most prominently used in the WKB approximation, especially if one can identify a magnitude gradation of the terms. This gives the Riccati equation for$u'$$$u''+u'^2+fu'+g=0.$$ In reverse, it sometimes helps to transform Riccati equations backwards this way to a second order linear equation. There is ... 0 The solutions (other than 0) don't seem to be expressible in closed form: at least, Maple can't find closed-form solutions. There are series solutions with coefficients$a_n$satisfying the recurrence: $$a_n + (-175 - 35 n) a_{n + 3} + (259 n^2 + 3367 n + 11008) a_{n + 6} + (-225 n^3 - 5400 n^2 - 42975 n - 113400) a_{n + 9} = 0$$ 1 $$p^2-py+x=0$$ This is D'Alembert's differential equation : $$y=xf(p)+g(p)$$ $$y=\dfrac x {p}+p$$ Differentiate both sides: $$p=\dfrac {p-xp'}{p^2}+p'$$ $$p^3-p=p'(p^2-x)$$ $$p(p^2-1)\dfrac {dx}{dp}=(p^2-x)$$ It's a first order linear DE. $$p(p^2-1)x'+x=p^2$$ 3 The +1 is time. Thus the independent variables considered together in, say, a 2+1 dimensional model live in a 3 dimensional space, but when considering time separate from space it makes sense to write it as 2+1, to emphasize that the spatial dimension is$2$but the dynamical problem is under investigation. I find this terminology is most common in ... 1 You need to correct the right-side function of your first-order system. It should read $$[y'(x),y''(x)] = F(x,[y(x),y'(x)])=\left[y'(x),\,-\frac{y'(x)^2}{y(x)}\right].$$ Then for the Lipschitz condition you get $$\|F(x,[u,v])-F(x,[a,b])\|=\left\|\left[v-b,\,-\frac{v^2}{u}+\frac{b^2}{a}\right]\right\| \le |v-b|+\frac1{|a|}(|v|+|b|)|v-b|+\frac{v^2}{|au|}|u-... 1$$ty'' +y'=0$$Rewrite the DE and integrate twice:$$(ty')'=0ty'=C_1y(t)=C_1 \color {red}{ \times \ln (t)}+C_2 \color {red}{\times 1}$$For the Wronskian you have that:$$W(\ln t,1)=\dfrac 1t \times 1 -0 \times \ln t=\dfrac 1 t$$The solution to this DE is the vector space (\ln t , 1) Note that for theorem of existence and uniqueness you need ... 1 So far you have$$ u(x, t) = \sum_n \sin\left(\frac{n\pi x}{l}\right)\left[ A_n e^{\lambda_+ t} + B_n e^{\lambda_- t} \right] \tag{1} $$where$$ \lambda_\pm = \frac{1}{2}\left[1 \pm \sqrt{1 - \left(\frac{2n\pi}{l}\right)^2}\right] $$Now consider the initial conditions u(x, 0) Replacing that in (1) you get$$ \sin x = \sum_n \sin\left(\frac{n\pi x}{l}\... 1 The first observation is that if$y$at some time during the numerical method takes a negative value, the exact solution will diverge towards$−∞$, and the numerical solution will follow. So what is observed is at least a move further away from the$x$-axis inside the integration interval$[1,2]. For the Euler method this excludes step sizes $\Delta x>0.... 0$(x-2y)~dx-(8x^2-9)~dy=0(x-2y)~dx=(8x^2-9)~dy(8x^2-9)\dfrac{dy}{dx}=x-2y\dfrac{dy}{dx}+\dfrac{2y}{8x^2-9}=\dfrac{x}{8x^2-9}$This is a first-order linear ODE. Of course follow the simplest properties. 3 Let$A=-a>0$so that $$I(t)=e^{-At}\int_0^te^{As}f(s)\,ds.$$ L'Hopital yields $$\lim_{t\to\infty}I(t)=\lim_{t\to\infty}\frac{\int_0^te^{As}f(s)\,ds}{e^{At}}=\lim_{t\to\infty}\frac{e^{At}f(t)}{Ae^{At}}=0$$ as$\lim\limits_{t\to\infty}f(t)=0\$.

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First of all, your differential equation is wrong. It should be $$x y'' + (1-x) y' + \lambda y = 0$$ In general, the solutions of this differential equation can be defined in terms of Kummer M and U functions: $$y \left( x \right) = c_1\,{{\mathrm M}\left(-\lambda,1,x \right)}+c_2 \,{{\mathrm U}\left(-\lambda,1,x\right)}$$ The Kummer M function is a ...

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