Please help (Can anyone do it step by step)?

Question

A particular solution of the differential equation
(d2y/dx2) + 4(dy/dx) + 4y = 8 sin2x

is y = p cos2x + q sin2x, where p and q are constants.

Find the value of p and q. Give that y = 1 and dy/dx = 0 at x = 0 find the solution of the differential equation.

Answer 1

The solution to this type of differential equation (linear 2nd order nonhomogeneous constant coefficient differential equation) is made up of 2 parts, the PARTICULAR solution and the HOMOGENEOUS solution.

To find the HOMOGENEOUS solution we solve the homogenous problem. The homogeneous problem is what you get when you replace all terms that DON'T have a y or a derivative of y with zero. In this case there's only 1 term that we replace with zero, the 8 sin2x term (all other terms have either a y or one of it's derivatives).

To find the homogeneous solution we need to solve
(d2y/dx2) + 4(dy/dx) + 4y = 0

The way to do this is to assume the solution y = e^(rx) and determine the values of r that satisfy the equation. Keep in mind that r is just a constant.

If y=e^(rx)
then dy/dx = r e^(rx) and d2y/dx2 = r^2 e^(rx)

Substituting into the homogeneous differential equation we get.

r^2 e^(rx) + 4r e^(rx) + 4 e^(rx) = 0

divide both sides of the equation by e^(rx) and we're left with
r^2 + 4r + 4 = 0

which has the repeated root r = -2 as the only solution.

Because the roots repeated the second solution gets an x in front of the exponential (This only happens for reapeated roots)

Thus the homogeneous solution is.
y = C1 e^(-2x) + C2 x e^(-2x)

WE DON'T DETERMINE THE VALUES OF C1 AND C2 YET! We do that when we have the whole solution (homogeneous and particular).

To find the PARTICULAR solution we substitute in the assumed form and determine the values of p and q that satisfy the original differential equation.

The problem says the particular solution is of the form
y = p cos(2x) + q sin(2x)

then
dy/dx = -2p sin(2x) + 2q cos(2x)
d2y/dx2 = -4p cos(2x) - 4q sin(2x)

substituting this into the differential equation gives us
[ -4p cos(2x) - 4q sin(2x) ] + 4[ -2p sin(2x) + 2q cos(2x) ] + 4[ p cos(2x) + q sin(2x) ] = 8 sin(2x)

multiplying out the right hand side and collecting like terms we get

8q cos(2x) - 8p sin(2x) = 8 sin(2x)

Now, we simply make sure the number of cos(2x) terms is the same on both sides of the equation and the number of sin(2x) terms is the same on both sides of the equation.

THE SIN(2x) TERMS
The left hand side has -8p sin(2x) and the right hand side has 8 sin(2x). In order for the 2 to be the same we need -8p = 8. This tells us that p = -1.

THE COS(2x) TERMS
The left hand side has 8q cos(2x) and the right hand side doesn't have any. In order for the the 2 sides to be the same we need 8q = 0. This tells us q = 0.

So the particular solution is
y = (-1) cos(2x) + (0) sin(2x)

or simply
y = - cos(2x).

We're almost done.

Now we worry about the intial conditions.

The whole solution is the sum of the homogeneous solution and the particular solution.

Thus our solution has the form
y = C1 e^(-2x) + C2 xe^(-2x) - cos(2x)

y=1 when x=0
1 = C1 e^(0) + C2 0 e^(0) - cos(0)
1 = C1 + 0 - 1
thus C1=2.

dy/dx=0 when x=0
dydx= - 2 C1 e^(-2x) + C2 e^(-2x) - 2 C2 xe^(-2x) + 2 sin(2x)
0 = - 2 (2) e^(0) + C2 e^(0) - 2 C2 0 e^(0) + 2 sin(0)
0 = - 4 + C2 - 0 + 0
thus C2 = 4.

Finally we arive at our solution,
y = 2 e^(-2x) + 4 xe^(-2x) - cos(2x)

Good Luck!

Answer 2

The equation is y" + 4y' +4y = 8 sin 2x. Given the particular solution, we want to choose p and q such that the particular solution satisfies the diff. eq. We find
y = p cos 2x + q sin 2x, y' = -2p sin 2x + 2q cos 2x, and y" = -4p cos 2x - 4q sin 2x. Thus. y" + 4y' + 4y = -8p sin 2x + 8q cos 2x. This is = 8 sin 2x when p = -1 and q = 0, so we have the particular solution

(1) y_p = - cos 2x.

For the general solution of the homogeneous equation, we have the auxilliary equation u^2 + 4u + 4 = 0, which has two equal roots u = -2. This tells us that we have the general solution y = Ae^(-2x) + Bxe^(-2x). We want this to be 1 when x = 0, so we find A = 1, and our solution is y = e^(-2x) + Bxe^(-2x). Since this should be 0 when x = 0, we see that B = 0. This gives us the solution

(2) y_g = e^(-2x).

Our complete solution is y_p + y_g.