Question

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Solve the PDE (D2 - DD') z = sin x cos 2 y​

Answer 1

Step-by-step explanation:

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Answer 2

To Find:

PDE(D2 - DD')z = sin x cos²y

Solution:

 PI = $\frac{1}{(D+3)(D+1)}$ $e^{e^{x} }$

     = $\frac{1}{D+3}$ $e^{-x}$ ∫$e^{e^{x} }$× $e^{x}dx$                     ∵$\frac{1}{D-a}$×$e^{ax}$∫$e^{-ax}$×dx

Putting $e^{x} = t$

         $e^{x}dx = dt$

       ∫$e^{e^{x} }$×$e^{x}dx =$∫$e^{t}dt=e^{t}=e^{e^{x} }$

     So, PI = $\frac{1}{D+3}e^{-x}$∫$e^{e^{x} }-e^{x}dx$

                = $\frac{1}{D+3}e^{-x}.e^{e^{x} }$

     Therefore,         P.I.  = $e^{-3x}$∫$e^{e^{x} }.e^{3x}.e^{-x}dx$