Question #291978

(D² +40D'² +3D' )Z=ycos x

1
Expert's answer
2022-01-31T16:31:50-0500

The actual question is;

(D2+4DD+3(D)2)z=ycosx,where D=x,and D=y\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 D=m,and D=1,\displaystyle D=m, \text{and }D^\prime=1, then the auxiliary equation of the given PDE in terms of variable m is;

m2+4m+3=0\displaystyle m^2+4m+3=0

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

Thus the complementary function of the given PDE is;

C.F=zc=f1(yx)+f2(y3x)\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;

P.I=1(D2+4DD+3(D)2)ycosx=1(D+D)(D+3D)ycosx\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

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

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

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

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

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

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

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

Thus the particular integral of the given PDE is;

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

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

z=f1(yx)+f2(y3x)ycosx+4sinx\displaystyle \color{red}{z=f_1(y-x)+f_2(y-3x)-y\cos x+4\sin x}


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