Answer to Question #225233 in Differential Equations for Shawon

Question

Solve the differential equation 2y^'''+ y^''- 8y^' - 4y = 7 - 2e^-x - cos² (2x).

Answer 1

Let

y=y_h+y_p

To find the homogeneous equitation,

An auxillary equation is

$2m^3+m^2-8m-4=0$

$2m^2(m+1)-4(2m+1)=0$

$(2m^2-4)(m+1)=0$

$(m+2)(m-2)(2m+1)=0$

$m=2,-2 ,-\frac 12$

Hence ,the homogeneous solution is

$y_h=C_1e^{\frac{-x}{2}}+C_2e^{-2x}+C_3e^{2x}$

To find the particular solution,

By trigonometric identities;

$7-2e^{-x}-cos^2(2x)=7-2e^{-x}-\frac12-\frac12cos(4x)=\frac{13}2-2e^{-x}-\frac12cos(4x)$

Now,

Let

$y_p=A+Be^{-x}+Ccos(4x)+Dsin(4x)$

By differentiation;

$y_p'=-Be^{-x}-4Csin(4x)+4Dsin(4x)$

$y_p''=Be^{-x}-16Ccos(4x)-16Dsin(4x)$

$y_p'''=-Be^{-x}+64Csin(4x)-64Dcos(4x)$

By substitution into the given equation;

$-2Be^{-x}+128Csin(4x)-128Dcos(4x)+Be^{-x}-16Ccos(4x)-16Dsin(4x)+8Be^{-x}+32Csin(4x)-32Dcos(4x)-4A-4Be^{-x}-4Ccos(4x)-4Dsin(4x)=\frac{13}2-2e^{-x}-\frac12cos(4x)$

Simplify;

$-4A+3Be^{-x}+(160C-20D)sin(4x)+(-160D-20C)cos(4x)=\frac{13}2-2e^{-x}-\frac12cos(4x)$

By comparison;

$-4A=\frac{13}2$

$A=-\frac{13}8$

$3B=-2$

$B=-\frac23$

$(160C-20D)=0$

$C=\frac18D$ ....(i)

Also,

$-160D-20C=-\frac12$ ...(ii)

Combining (i) and (ii) ,

$D=\frac {1}{325}$

$C=\frac{1}{2600}$

Substitute the constants into y_p;

$y_p=-\frac{13}8-\frac23e^{-x}+\frac1{2600}cos(4x)+\frac1{325}sin(4x)$

Hence the general solution of the question is

$y=y_h+y_p$

$y=C_1e^{-\frac x2}+C_2e^{-2x}+C_3e^{2x}-\frac{13}2-\frac23e^{-x}+\frac{1}{2600}cos(4x)+\frac1{325}sin(4x)$