Question

solve (D³ – 7DD'² –6D'³)z = sin(x+2y) + e^2x+y

Answer 1

$$Solution$$

$m^3-7m =0;\ m(m^2-7)=0\\ \therefore m_1=0,m_2=\sqrt{7}\\ C.F=\phi_1 (y+m_1 \cdot x)+\phi_2 (y+m_2 \cdot x)+x\ \phi_3 (y+m_2 \cdot x)\\ C.F=\phi_1 (y+0 \cdot x)+\phi_2 (y+(\sqrt{7}) \cdot x)+x\ \phi_3 (y+(\sqrt{7}) \cdot x)\\$

$P.I=\frac{sin(x+2y) + e^{2x}+y}{D³ – 7DD'² –6D'³}\\ F(D,D')=D³ – 7DD'² –6D'³;\ Where\ a=1, b=2\\ f(a,b)=(1,2)=1^3-7(1)(2)^2-6(2)^3=1-7(4)-6(8)\\ \implies-75 \ne0\\$

$P.I=\frac{1}{f(a,b)}\intop \intop sin\ v\ \delta v \delta v + \intop \intop (e^{2x}+y)\ \delta y \delta x \\ \implies \frac11\intop \intop -cos\ v\ \delta v \delta v \\ \implies 1\intop \intop -sin\ v\ \delta v =cot\ v= cot(x+2y)\\ and\\ \implies \intop \intop (e^{2x}+y)\ \delta y \delta x =\intop e^{2x}+y)\ \delta y = e^{2x}y+\frac{y^2}{2}\\ \implies \frac{y^2}{2}x+y \frac12 e^{2x}\\ P.I =cot(x+2y)+\frac{y^2}{2}x+y \frac12 e^{2x}\\$

General Solution

$C.F+P.I= \phi_1 (y+0 \cdot x)+\phi_2 (y+(\sqrt{7}) \cdot x)+x\ \phi_3 (y+(\sqrt{7}) \cdot x)+cot(x+2y)+\frac{y^2}{2}x+y \frac12 e^{2x}\\$