Answer to Question #199333 in Differential Equations for Bharat

Question

(D²-D'²-3D+3D')z=xy+7

Answer 1

Solution:-

$(D^2-D'^2-3D+3D')z=xy+7$

$(D-D')(D+D'-3)z=xy+7$

$m_1=1 \\ m_2=-1 \\ c_1=0 \\ c_2=3$

$c.f=e^{c_1x}f_1(y+m_1x)+e^{cx}f_2(y+m_2x)$

$c.f=e^{0x}f_1(y+x)+e^{3x}f_2(y-x)$

$c.f=f_1(y+x)+e^{3x}f_2(y-x)$

PI=$\frac{xy+7}{(D-D')(D+D'-3)}$

$D'=-1\\ D=1$

$PI = \frac{xy+7}{-6}$

answer = C.F + P.I

=$f_1(y+x)+e^{3x}f_2(y-x) - (\frac{xy+7}{6})$