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I have the following : $$ u_{xx}-u_{xy}+u_{y}-u=cos(x+2y)+e^y : u=u(x,y)$$ the par1ticular solutuion for the part $ e^y$ : $$\frac{1}{D_{1}^2-D_{1}D_{2}+D_{2}-1}e^y=\frac{1}{(D_{1}-1)(D_{1}-D_{2}+1)}e^y$$ I had to deal first with : $$(D_{1}-D_{2}-1)u =e^y$$ then dealing with $$\frac{1}{D_{1}-1}u $$ I wonder if there is a short solution which I haven't noticed ?

Thank you ...

Mahmoud Hassan
  • 1,132

1 Answers1

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let $$ u=\frac{1}{(D_{1}-1)(D_{1}-D_{2}+1)}e^y$$ which implies $$u_{xx}-u_{xy}+u_{y}-u=e^y \ (1)$$ let $ u(x,y)=f(x)e^y$ So we have : $$f''(x)e^y-f'(x)e^y+f(x)e^y-f(x)e^y=e^y \Rightarrow f''(x)-f'(x)-1=0 $$ which implies: $$ f(x) =-x $$ finally $$u =-xe^y $$

Mahmoud Hassan
  • 1,132