Question

solve the partial differential equation (2D²-5D+2D` ²)z=5sin(2x+y)​

Answer 1

Answer:

Find an answer to your question solve the partial differential equation ( 2D²-5D+2D` ²)z=5sin(2x+y)

Answer 2

Answer:

$z=f_{1}(y+2x)+f_{2}(y+\frac{1}{2}x)-\frac{5}{3}xcos(2x+y)$

Step-by-step explanation:

Step 1 of 2

The auxiliary equation is formed as:

$2m^{2}-5m+2=0\\(2m-1)(m-1)=0\\m=2\\m=\frac{1}{2}$

Step 2 of 2:

$C.F = f_{1}(y+2x) +f_{2}(y+\frac{1}{2} x)$

$PI = \frac{1}{2D^{2}-5DD^{'} +2D^{r2} } .5sin(2x+y)\\PI =I.P. \frac{1}{ (2D-D^{'})(D-2D^{'}) } .5e^{i(2x+y)}$

$PI = IP\frac{1}{((2)(2i)-i)} . 5x e^{i(2x+y)}\\PI = IP(\frac{-i}{3}(5x)[cos(2x+y)+isin(2x+y)] \\PI = (\frac{-5}{3}x)cos(2x+y)$

So we get the final answer as :

$z=C.F+PI\\z=f_{1}(y+2x)+f_{2}(y+\frac{1}{2}x)-\frac{5}{3}xcos(2x+y)$