# Thread: partial differentials: ∂∂tA(y,t)+6ΛΩ(y2−y)sin(t)=∂2∂y2A(y,t)

1. ## partial differentials: ∂∂tA(y,t)+6ΛΩ(y2−y)sin(t)=∂2∂y2A(y,t)

Problem
${\frac {\partial }{\partial t}}A\left( y,t \right) +6\,\Lambda\,\Omega\, \left( {y}^{2}-y \right) \sin \left( t \right) ={\frac {\partial ^{2}}{\partial {y}^{2}}}A \left( y,t \right)$

${\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0$

Boundary condition

${\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0$

I don't know how to get this kind of solution. this is my attempt . please help.

2. Originally Posted by Grindra
Problem
${\frac {\partial }{\partial t}}A\left( y,t \right) +6\,\Lambda\,\Omega\, \left( {y}^{2}-y \right) \sin \left( t \right) ={\frac {\partial ^{2}}{\partial {y}^{2}}}A \left( y,t \right)$

${\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0$

Boundary condition

${\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0$

I don't know how to get this kind of solution. this is my attempt . please help.
The first place you might start is the separation of variables method, see for example https://math.stackexchange.com/quest...-clarification
...
Assume we have an inhomogeneous Partial Differential Equation of the form Auxx+2Buxy+Cuyy+Dux+Euy+Fu=w(x,y)(1)

with some initial and boundary conditions.
Let us define theauxiliary linear homogeneous equation as
Avxx+2Bvxy+Cvyy+Dvx+Evy+Fv=0(2)
with the same boundary conditions as in (1) Then the general solution u=ugeneral
of the inhomogeneous equation (1) can be written as the sum of a particular solution uparticular of (1) and a general solution vgeneral of the auxiliary homogeneous equation(2):
ugeneral(x,y)=vgeneral(x,y)+uparticular(x,y).

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