Answer to Question #139807 in Differential Equations for Nikhil Singh

Question

Solve
(D^2+DD'-6D'^2)z= ycosx

Answer 1

Given,

$$(D^2+DD'-6D'^2)z= y\cos x$$

Now, Auxiliary equation will be

$$m^2+m-6=0\\ \implies (m-2)(m+3)=0\\ \implies m=2,-3$$

Hence,

$$C.F=\phi_1(y+2x)+\phi_2(y-3x)$$

Now,

$$P.I=\frac{1}{(D^2+DD'-6D'^2)}y\cos x\\ \implies P.I=\frac{1}{(D+3D')(D-2D')}y\cos x\\ =\frac{1}{D+3D'}\int_{D-2D'}y\cos x \, dx\\ =\frac{1}{D+3D'}[(c_1-2x)\sin x-2\cos x]\\ =\frac{1}{D+3D'}(y\sin x-2\cos x)\\ =\int_{D+3D'}y\sin xdx-\int_{D+3D'}2\cos x dx\\ =\int_{D+3D'}(c_2+3x)\sin xdx-\int_{D+3D'}2\cos x dx\hspace{1cm}(y=c_2+3x)\\$$

$$\implies P.I=-y\cos x+\int\cos x \, dx =-y\cos x+\sin x$$

Therefore the complete solution is

$C.F+P.I=\phi_1(y+2x)+\phi_2(y-3x)+-y\cos x+\sin x$