Question

Solve the partial differential equation (D ^3 − 7DD'^2 − 6D '^3 )z = e ^(2x+y)​

Answer 1

Answer:

OK that's very easy answer 9x+y

Answer 2

Answer:

The solution of the given partial differential equation is given by  $z=f_{1} (y-x)+f_{2} (y-2x)+f_{3} (y+3x)-\frac{1}{12}e^{2x+y}$

Step-by-step explanation:

The Partial differential equation is

$(D^{3}-7D(D^{1}) ^{2}-6 (D^{1} )^{3})z=e^{2x+y}$

Replacing D with m and $D^{1}$ by 1, we get an Auxilary equation

$m^{3}-7m-6=0$

The roots are m = -1, -2, 3

Therefore the Complimentary function is given by

$f_{1} (y-x)+f_{2} (y-2x)+f_{3} (y+3x)$

Now, Particular integral = $\frac{1}{D^{3}-7D(D^{1}) ^{2}-6 (D^{1} )^{3}} e^{2x+y}$

Replace D with 2 (coefficient of x) and $D^{1}$ with 1(coefficient of 1) we get

P. I = $\frac{1}{2^{3}-7(2)(1)-6(1 )^{3} } e^{2x+y}$

P. I =  $\frac{1}{-12}e^{2x+y}$

Complete solution = C.F + P.I

$z=f_{1} (y-x)+f_{2} (y-2x)+f_{3} (y+3x)-\frac{1}{12}e^{2x+y}$