Answer to Question #291978 in Differential Equations for Omprakash

Question

(D² +40D'² +3D' )Z=ycos x

Answer 1

The actual question is;

$\displaystyle (D^2+4DD^\prime+3(D^\prime)^2)z=y\cos x, \text{where }D=\frac{\partial}{\partial x}, \text{and}\ D^\prime =\frac{\partial}{\partial y}$

To find the complementary function (C.F), let $\displaystyle D=m, \text{and }D^\prime=1,$ then the auxiliary equation of the given PDE in terms of variable m is;

$\displaystyle m^2+4m+3=0$

$\displaystyle \Rightarrow (m+1)(m+3)=0\Rightarrow m=-1,-3.$

Thus the complementary function of the given PDE is;

$\displaystyle \color{green}{C.F=z_c=f_1(y-x)+f_2(y-3x)}$

Next, to find the particular integral (P.I) of the given PDE, we proceed as follows;

$\displaystyle P.I=\frac{1}{(D^2+4DD^\prime+3(D^\prime)^2)}y\cos x=\frac{1}{(D+D^\prime)(D+3D^\prime)}y\cos x$

$\displaystyle \ =\frac{1}{(D+D^\prime)}\int (c+3x)\cos x\ dx, \text{where }y=c+3x\\$

$\displaystyle \ =\frac{1}{(D+D^\prime)}[(c+3x)\sin x+3\cos x], \text{ by integration by parts}$

$\displaystyle \ =\frac{1}{(D+D^\prime)}[y\sin x+3\cos x], \text{ since }y=c+3x$

$\displaystyle \ =\int[(c+x)\sin x+3\cos x]\ dx, \text{where we put }y=c+x$

$\displaystyle \ =\int[(c+x)\sin x\ dx]+3\int \cos x\ dx$

$\displaystyle \ =[(c+x)(-\cos x)+\sin x]+3\sin x\ ,\text{integration by parts}$

$\displaystyle \ =-y\cos x+4\sin x,\ \text{where we put back }y=c+x$

Thus the particular integral of the given PDE is;

$\displaystyle \color{blue}{P.I=z_p=-y\cos x+4\sin x}$

Since complete solution (z) is $\displaystyle \color{red}{z=z_c+z_p}$, thus we have the complete solution as;

$\displaystyle \color{red}{z=f_1(y-x)+f_2(y-3x)-y\cos x+4\sin x}$