Answer:
a.P.I=[tex]\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)[/tex]
b.G.S=[tex]C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x}[/tex]
Step-by-step explanation:
We are given that a linear differential equation
[tex]y''+2y=e^{3x}+x^3[/tex]
We have to find the particular solution
P.I=[tex]\frac{e^{3x}}{D^2+2}+\frac{x^3}{D^2+2}[/tex]
P.I=[tex]\frac{e^{3x}}{3^2+2}+\frac{1}{2} x^3(1+\frac{D^2}{2})^{-2}[/tex]
P.I=[tex]\frac{e^{3x}}{11}+\frac{1-2\frac{D^2}{4}+3\frac{D^4}{16}+...}{2}x^3[/tex]
P.I=[tex]\frac{e^{3x}}{11}+\frac{x^3-2\frac{\cdot3\cdot 2x}{4}}{2}+0}[/tex] (higher order terms can be neglected
P.I=[tex]\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)[/tex]
b.Characteristics equation
[tex]D^2+2=0[/tex]
[tex]D=\pm\sqrt2 i[/tex]
C.F=[tex]C_1cos \sqrt2x+C_2 sin\sqrt2 x[/tex]
G.S=C.F+P.I
G.S=[tex]C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x)[/tex]
