2nd order differential equation

[james.farrow]

I have the following equation

d^2y/dx^2 +4dy/dx +8y = 4sin(2x) - 12cos(2x) y(0)=1 & y'(0)=-6

For the auxillary equation I have (m + 2)^2 = sqrt -4

which gives m=-2-2i & m=-2+2i

Which gives y=e^-2x(Ccos(-2x) + Dsin(-2x))

Now to tackle particular integral.

Try y=acos(2x) + bsin(2x)

After differentiatiing twice and equating I have a=-5 and b=-9


Finally I get y=e^-2x(Ccos(-2x) + Dsin(-2x)) - 5cos(2x) -9sin(2x)

After aplying the initial conditions my final answer is

y=6e^-2xcos(-2x) - 18e^-2sin(-2x) - 5cos(2x) - 9sin(2x)


In all my working it is only e that is raised to the power of -2x, I know it looks like the whole statement is raised but it isnt, the e^-2x is multiplied by the trig expression.

James

 

[Mark44]

What's your question? You can check your solution yourself by confirming that for your solution,
1. y(0) = 1 and y'(0) = -6, and
2. y'' + 4y' + 8y = 4sin(2x) - 12cos(2x)

 

[james.farrow]

That's just it! I've tried putting my solution back in n it doesn't work... I'm going to have another go at it but I can't see where I've gone wrong!

 

[Dick]

Your particular solution -5cos(2x)-9sin(2x) is wrong. It doesn't solve y''+4y'+8y=4*sin(2x)-12*cos(2x). There's something wrong with how you got 'a=-5 and b=-9'.

 

[Mark44]

You made a mistake on you calculation of the coefficients of your particular solution. I get a = -1, b = -1.

Also, the solution to the homogeneous equation should be y_p = e^-2x(C*cos(2x) + D*sin(2x)). You have cos(-2x) and sin(-2x).