Recent content by kosovtsov

The general solution to your PDE is q(x,y,t) =C(x,axt+y)\exp(\frac{xy}{a}+\frac{y^3}{3ax}), where C(x,y) is an arbitrary function. So q(x,y,t) =δ(x,axt+y)\exp(-\frac{t(a^2t^2x^2+3atxy+3x^2+3y^2)}{3}) is the solution with q(x,y,0) =δ(x,y).

The general solution to your ODE is as follows y(x) = -3x(x-3)\ln(x)+[(-\frac{1}{2}x^2+x)\int_{-x}^∞ \frac{\exp(-t)}{t}dt -\frac{1}{2}\exp(x)(x-1)]C_1-\frac{x^3+9}{2}+x^2C_2+xC_3 where C_i are arbitrary constants.

Unfortunately, I can not grasp the physical sense of your problem out of hand from your picture, but it seems to me that you may analyse the analytical solution of your problem. For linear A(x)=ax+b the general solution to your PDE is relatively easy: u(t,x) =...

The main idea: If the homogeneous DE \alpha(x,y)\frac{\partial s(x,y)}{\partial x}+\beta(x,y)\frac{\partial s(x,y)}{\partial x}=0 is solvable (it is sufficient to find any particular solution), then DE of the following type \alpha(x,y)\frac{\partial p(x,y)}{\partial...

The general solution to your PDE is u(x,y) = -1-x+e^x F(x^2-2y), where F is an arbitrary function. Your boundary condition leads to F(t) = (1+t^{1/2}+t)exp(-t^{1/2}), where t=x^2-2y.

Maple solves your ODE via Legendre functoins.

Unfortunately I can not realize your goal here. If you interested in steady state conditions, you have to consider the right equations from the top three expressions + the fourth and try to solve this algebraic system. It’s really easy! (There is misprint in fourth expression! Examine...

It seems to me that something is wrong here. Your B(t)=tanh(t); A(t)=i.sech(t) is not a solution to your system when s=0. Or I misunderstand your notations.

I'm sorry for misprint. The right answer is w(x,y)=\int_{-\infty}^{-\infty} \{F_1 (\omega) AiryAi [-((A\omega^2+C)/B)^{1/3}x]+F_2 (\omega) AiryBi [-((A\omega^2+C)/B)^{1/3}x]\}\exp(y\omega)d\omega ,

Your PDE A\frac{\partial^2 w(x,y)}{\partial y^2}+\frac{B}{x}\frac{\partial^2 w(x,y)}{\partial y^2}+Cw(x,y) = 0 can be solved by the Laplace transform. The general solution is as follows w(x,y)=\int_{-\infty}^{-\infty} F_1 (\omega) AiryAi [-((A\omega^2+C)/B)^{1/3}x]+F_2 (\omega) AiryBi...

Your PDE can be solved with help of Laplace transform method. For your purpose it'll be better the following form of general solution ( I assume that in fact R is r) x(t,r) = \frac{1}{r}[\int_c^tf(\xi)g(vt-v\xi+r)(vt-v\xi+r)d\xi+F(vt+r)], where F(z) is an arbitrary function, c is an...

You get \frac{d u}{dx} = 1+\sin(u) and then \int\frac{d u}{1+\sin(u)} = x+C, where C is an arbitrary constant, or -\frac{2}{\tan[\frac{u}{2}]+1}= x+C so the general solution to your ODE is y(x) = -2\arctan(\frac{2+x+C}{x+C})-x-3 Substituting x=0 you find that C=-2, so...

Can you give an example, please?

Your ODE is not an ODE with constant coefficients, so it can not be solved by Laplace Transform. An integrating factor to your ODE \mu=t\exp(-\frac{t^2}{2}) and corresponding first integral is I=\exp(-\frac{t^2}{2})(t\frac {d y}{d t}-y+1)+C1 in sense that \frac {d I}{d...

The equivalent splited system is as follows q= (\frac {\partial^2 y}{\partial t^2}+A\frac {\partial y}{\partial t}-B\frac {\partial ^2 y}{\partial z^2})/C \frac {\partial ^4y}{\partial t^4}= B\frac {\partial ^4y}{\partial t^2 \partial z^2}-A\frac {\partial ^3y}{\partial t^3}-AD\frac...