Question #120992
sove pde d²z/dx²+d²z/dxdy+dz/dy-Z=e^-x
1
Expert's answer
2020-06-09T18:28:42-0400

2z2x2+2zxy+zyz=ex\frac{\partial^{2}z}{\partial^{2}x^{2}} + \frac{\partial^{2}z}{\partial x\partial y} + \frac{\partial z}{\partial y} -z = e^{-x}


Equation can be written as:


(D2+DD+D1)z=ex(D^{2} + DD' + D' - 1)z = e^{-x}


(D+1)(D+D1)z=ex(D + 1)(D + D' -1)z = e^{-x}

Comparing above equation with

(Dm1Dα1)(Dm2Dα2)=F(x,y)(D -m_1 D' - \alpha_1)(D -m_2 D' - \alpha_2) = F(x,y)

so m1=0,m2=1,α1=1,α2=2m_1 = 0, m_2 = -1, \alpha_1 = -1, \alpha_2 = 2


So C.F. of the equation is

C.F.=exϕ1(y)+exϕ2(yx)C.F. = e^{-x}\phi_1(y) + e^{x}\phi_2(y-x)


then PI is calculated as

P.I.=1(D+1)(D+D1)ex=1(D+1)(D1)exP.I. = \frac{1}{(D+1)(D+D'-1)}e^{-x} = \frac{1}{(D+1)(D-1)}e^{-x} as exponential has no y.


P.I.=1(D+1)(D1)exP.I. = \frac{1}{(D+1)(D-1)}e^{-x} =ex1D(D2)1= e^{-x} \frac{1}{D(D-2)}1 =xex2= -\frac{xe^{-x}}{2}


So solution to the problem is

z(x,y)=exϕ1(y)+exϕ2(yx)xex2z(x,y) = e^{-x}\phi_1(y) + e^{x}\phi_2(y-x) - \frac{xe^{-x}}{2}


Solution can be verified by taking partial derivatives and put them in differential equation.


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