Question

Solve the following initial value problem d^2 y/dx^2 +dy/dx -2y =-6 sin 2x -18 cos 2x , y(0) = 2, y'(0) = 2

Answer 1

The given equation can be written as

(D²+D-2)y = -6 sin2x -18cos2x

The auxiliary equation of the above is

D² +D-2=0

D = 1,-2

The complimentary solution is given by:

y= c₁ e^x + c₂ e^-2x

Particular solution is given by:

$\frac{1}{D^2+D-2}( -6sin2x -18cos2x )=$

$=\frac{1}{D^2+D-2} (-6sin2x)-\frac{1}{D^2+D-2}(-18cos2x)$

$=\frac{-6sin2x}{D-6} - \frac{18cos2x}{D-6}$

$= (D+6) \frac{-6sin2x}{D^2-36}-(D+6) \frac{18cos2x}{D^2-36}$

$=(D+6) \frac{-6sin2x}{-40} -(D+6) \frac{18cos2x}{-40}$

$=\frac{3}{10}(cos2x+3sin2x) + \frac{9}{10} (-sin2x+3cos2x)$

=3 cos2x

So, the solution is y= c₁ e^x+ c₂e^-2x + 3 cos2x

Now, initial conditions given are:

y(0)=2

-1=c₁+c₂

y' = c₁e^x -2c₂e^-2x -6 sin2x

y'(0)=2

2= c₁-2c₂

c₁=0 , c₂=-1

y=-e^-2x +3cos2x