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# Revision history [back]

Here is what I get.

s^2*laplace(y(t, x), t, s) - D[0, 0](laplace)(y(t, x), t, s)*D(y)(t, x)^2 - s*y(0, x) - D(laplace)(y(t, x), t, s)*D[1, 1](y)(t, x) - D(y)(0, x) == k*sin(pi*x)/s


Within this expression, the meanings are:

$\frac{\partial y}{\partial t}$ = D(y)(t, x)

$\frac{\partial y}{\partial x}$ = D(y)(t, x)

$\frac{\partial^2 y}{\partial t^2}$ = D[0, 0](y)(t, x)

$\frac{\partial^2 y}{\partial x^2}$ = D[1, 1](y)(t, x)

So, D(y)(0, x) is $\frac{\partial y}{\partial t}|_{t=0}$.

You can check the first one above by doing diff(y,t).

Here is what I get.

s^2*laplace(y(t, x), var('y t s k')
y=function('y',x,t)
de=diff(y,t,2)-diff(y,x,2)==k*sin(pi*x)
de.laplace(t,s)


gives

s^2*laplace(y(x, t), t, s) - D[0, 0](laplace)(y(t, x), D(y)(x, t)^2*D[0, 0](laplace)(y(x, t), t, s)*D(y)(t, x)^2 s) - s*y(0, x) s*y(x, 0) - D(laplace)(y(t, x), D(laplace)(y(x, t), t, s)*D[1, 1](y)(t, x) s)*D[0, 0](y)(x, t) - D(y)(0, x) D(y)(x, 0) == k*sin(pi*x)/s


Within this expression, the meanings are:

$\frac{\partial y}{\partial t}$ = D(y)(t, x)D(y)(x, t)

$\frac{\partial y}{\partial x}$ = D(y)(t, x)D(y)(x, t)

$\frac{\partial^2 y}{\partial t^2}$ = D[0, 0](y)(t, x)D[1, 1](y)(x, t)

$\frac{\partial^2 y}{\partial x^2}$ = D[1, 1](y)(t, x)D[0, 0](y)(x, t)

So, D(y)(0, x) is $\frac{\partial y}{\partial t}|_{t=0}$.

You can check the first one above by doing diff(y,t).