d^2y/dx^2 + 2 dy/dx +5y=0
Very confused by how to do this, I have found the aux equation and it’s roots (-1+-2i) and put the equation into the form:
y=e^ax( C cos bx + D sin bx)
But now I need to find out the solution when y=1, x=0 and dy/dx=1.
But I’m not sure what part I am differentiating. Can anyone guide me a little?
y" + 2y' + 5y = 0
The roots of D^2+2D+6=0 are -1±2i, so the solution to the DE is
y = e^-x (a sin2x + b cos2x)
y' = e^-x ((2a-b)cos2x - (a+2b)sin2x)
Now plug in your values to solve for a and b
Thanks oobleck,
Having done this if it seems that D, C are 0 or 1 then that means the explicit solution is just the same as the solution found just without D and C?
I arrived at 1=-1C+D+1+2D, am I totally of track?
To find the solution to the given equation when y = 1, x = 0, and dy/dx = 1, let's go through the steps together:
1. Start with the general solution of the differential equation, which you already have:
y = e^ax(C cos(bx) + D sin(bx))
2. Now, we need to determine the values of a, b, C, and D using the given aux equation roots and the initial conditions.
3. The auxiliary equation for the given differential equation is:
r^2 + 2r + 5 = 0
Solve this quadratic equation by factoring or using the quadratic formula. Since you already mentioned that the roots are -1 ± 2i, we'll assume you have the correct root values.
4. From the roots, we can determine the values of a and b:
a = -1
b = 2
5. Using the initial condition y = 1, x = 0, we substitute these values into the general solution:
1 = e^(a*0)(C cos(b*0) + D sin(b*0))
Simplifying this further, we get:
1 = C
So, C = 1.
6. Using the initial condition dy/dx = 1, x = 0, we substitute these values into the derivative of the general solution:
1 = a e^(a*0)(C cos(b*0) + D sin(b*0)) + b e^(a*0)(-C sin(b*0) + D cos(b*0))
Simplifying this further, we get:
1 = a C + b D
Since we know that C = 1, we can rewrite the equation as:
1 = a + b D
Now, solving for D, we have:
D = (1 - a) / b
Substituting the values of a = -1, b = 2, we find:
D = (1 - (-1)) / 2
D = 1
7. Now that we have the values of C and D, we can write the final solution:
y = e^(-x)(1 cos(2x) + 1 sin(2x))
Simplifying this gives:
y = e^(-x)(cos(2x) + sin(2x))
So, the solution to the given differential equation, when y = 1, x = 0, and dy/dx = 1, is y = e^(-x)(cos(2x) + sin(2x)).