Differential Equations question... PLEASE help.?

Question

I'm trying to find the general solution to 6y''+4y'+4y=0.

I know I find the roots, which are (-4 +sqrt(80)i)/12 and (-4-sqrt(80)i)/12. I know you have to plug them into the equation y=c1y2+c2y2... but how do i get y1 and y2?

since it's imaginary, i know it's something to do with sin and cos but i'm not sure.. please help!

Answer 1

For these kind of linear homogenous constant coefficient differential equation, the solution are always of the for y=e^mx.

So we assume that and then we plug in y=e^mx, we get the equation (6m^2+4m+4)e^mx=0

Divide by e^mx because it will never be zero. Then we have a quadratic and we can solve for m.

As you said,
m1=(-1+i*sqrt(5))/3
m2=(-1-i*sqrt(5))/3
after we simplify the expression

So you simply plug them into y=e^mx. Since the m1 and m2 are different, you have found linearly independent slutions to your differential equation and your general solution will be the linear combination of these two solutions.

But now the problem is that you have a complex expression.
y1=e^((-1+i*sqrt(5))/3)x
y2=e^((-1-i*sqrt(5))/3)x

But we started out with real coefficients and we must have 2-real values linearly independent solutions.

So, we use a (very) handy little thing given to us by Euler
e^(i*x)=cos(x)+i*sin(x)

Notice that y1 and y2 can be broken into their real and imaginary parts like this

y1=e^(-x/3)(e^(i*sqrt(5)/3))
y2=e^(-x/3)(e^(-i*sqrt(5)/3))

Using Euler's formula we get two linearly independent solutions of
y1=(e^-x/3)(cos(x*sqrt(5)/3))
y2=(e^-x/3)(sin(x*sqrt(5)/3))

Take their linear combination and we have the general solution
y=c1*(e^-x/3)(cos(x*sqrt(5)/3)) + c2*(e^-x/3)(sin(x*sqrt(5)/3))

Basically, if you get complex roots, e to the power of the real part and sine and cosine with the imaginary part will give you the solution.

Answer 2

dy/dx = (x^2)(8 + y) First, we separate the variables by way of multiplying by way of dx and dividing by way of (8 + y): dy/(8 + y) = (x^2)dx combine the two sides, remembering the left will use a organic log and the perfect could have a relentless: ln(8 + y) = (a million/3)x^3 + C develop e to the flexibility of the two section: 8 + y = Ae^((a million/3)x^3) Subtract 8 and we've the final form: y = Ae^((a million/3)x^3) - 8 Now, we've the factor (0, 3), so plug those in to discover a and the particular answer: 3 = Ae^((a million/3)(0)^3) - 8 resolve for A: 11 = Ae^((a million/3)(0)) 11 = Ae^(0) A = 11 So, the particular answer is: y = 11e^((a million/3)x^3) - 8

Answer 3

Hi Katie,

Use the method of the varyoing constants :
y=c1y2+c2y2
differentiate this equation twice and equals the 6 to the constants before your y'' , same with y' and y

Answer 4

best thing i can think of is take 4y to the other side thereby giving

6ydoubleprime +4yprime= -4y

factor one prime out thereby

y'(6y'+4)=-4y

y'= -4y/6y'+4

now you should have been told either what y''= or y'= because now you can't really continue the question without these key information