Answer to Question #218174 in Differential Equations for Dhanush

Question

Solve (D2
-3D+2)y = 2cos (2x+3)+2 ex
.

Answer 1

Solution;

(D²-3D+2)y=2cos(2x+3)+2e^x

Rewrite;

D²y-3Dy+2y=2cos(2x+3)+2e^x

y"-3y'+2y=2cos(2x+3)+2e^x

The equation can be solved using variation of parameters.

The general solution takes the form ;

y=y_c+y_p

To solve for y_c;

An auxiliary equation can be given as;

m²-3m+2=(m-2)(m-1);

m=2 or m=1

y_c=C₁e^2x+C₂e^x

To solve y_p let y_p=y_p1+y_p2 since g(x) of the equation is a combination.

Solve y_p1 from y"-3y'+2y=2cos(2x+3)

Let y_p1=Acos(2x+3)+Bsin(2x+3)

$y'_{p1}$=-2Asin(2x+3)+2Bsin(2x+3)

$y''_{p1}$ =-4Acos(2x+3)-4Bcos(2x+3)

Replace back in the equation;

-4Acos(2x+3)-4Bsin(2x+3)-3{-2Asin(2x+3)+2Bcos(2x+3)}+2{Acos(2x+3)+Bsin(2x+3)}=2cos(2x+3)

Distribute;

(-4A-6B+2A)cos(2x+3)+(-4B+6A+2B)sin(2x+3)=2cos(2x+3)

This means;

-4A-6B+2A=2

-4B+6A+2B=0

Solve for A and B using both equation;

B=3A

-4A-6×3A+2A=2

-4A-18A+3A=2

A=$\frac{-1}{10}$ ,B=-$\frac{3}{10}$

So that;

y_p1=$\frac{-1}{10}$ cos(2x+3)-$\frac{3}{10}$ sin(2x+3)

Solve y_p2 from y"-3y'+2y=2e^x

Let y_p2=Axe^x

$y'_{p2}$ =A(e^x+xe^x)e^x=A(1+x)e^x

$y''_{p2}$ =A(2e^x+xe^x)=A(2+x)e^x

Substitute in the equation;

A(2+x)e^x-3(A(1+x)e^x)+2Axe^x=2e^x

Distribute;

(2A+Ax-3A-3Ax+2Ax)e^x=2e^x

2A-3A=2

A=-2

y_p2=-2xe^x

The general solution of the equation will be;

y=y_c+y_p

y=C₁e^2x+C₂e^x-$\frac{1}{10}$cos(2x+3)-$\frac{3}{10}$sin(2x+3) -2xe^x