∂2x2∂2z+∂x∂y∂2z+∂y∂z−z=e−x
Equation can be written as:
(D2+DD′+D′−1)z=e−x
(D+1)(D+D′−1)z=e−x
Comparing above equation with
(D−m1D′−α1)(D−m2D′−α2)=F(x,y)
so m1=0,m2=−1,α1=−1,α2=2
So C.F. of the equation is
C.F.=e−xϕ1(y)+exϕ2(y−x)
then PI is calculated as
P.I.=(D+1)(D+D′−1)1e−x=(D+1)(D−1)1e−x as exponential has no y.
P.I.=(D+1)(D−1)1e−x =e−xD(D−2)11 =−2xe−x
So solution to the problem is
z(x,y)=e−xϕ1(y)+exϕ2(y−x)−2xe−x
Solution can be verified by taking partial derivatives and put them in differential equation.
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