Answer to Question #311594 in Differential Equations for Tria

Question

(D^2+4D+3)y=2cos^2x

Answer 1

The general solution takes the form

y=y_c+y_p

To solve for y_c;

An auxiliary equation can be given as:

$m^2+4m+3=0$

D=16-12=4

$m_1=\frac{-4+2}{2}=-1$

$m_2=\frac{-4-2}{2}=-3$

$y_c=C_1e^{-x}+C_2e^{-3x}$

2cos²x=cos(2x)+1

$y_p=x^se^{ax}(R_m(x)cos(bx)+T_m(x)sin(bx)($

s=0 if a+bi is not a root or s=k if it is.

Sollutiong for 1:

a+bi=0, then s=0

y₀=A

y'₀=0

y''₀=0

3A=1

A=1/3

y₀=1/3

Sollution fot cos(2x)

a+bi=2i, then s=0

y₁=Bsin(2x)+Acos(2x)

y'₁=2Bcos(2x)-2Asin(2x)

y''₁=-4Bsin(2x)-4Acos(2x)

(-B-8A)sin(2x)+(8B-A)cos(2x)=cos(2x)

A=-1/65

B=8/65

$y_1=\frac{8sin(2x)}{65}-\frac{cos(2x)}{65}$

$y=C_1e^{-x}+C_2e^{-3x}+ \frac{8sin(2x)}{65}-\frac{cos(2x)}{65}+1/3$