Question

y"-2y'+y=cos2t

1. general solution of corresponding homongenous equation

2. particular solution

3.solution of initial value problem with initial conditions y(0)=y'(0)=0

Answer 1

Undetermined coefficeients method

We can solve the given initial value problem by using method of variation of parameters method.

So the most general solution is   Y = Y_C + Y_P

Y_C = Homogeneous solutin

Y_P = Particular solution.

Now the given initil value problem is y’’-2y’+y = cos(2t)

So let us take the homogeneous equation for solving

So y’’-2y’+y = 0    -------------------(1)

Now let us find the characteristic equation for homogeneous equation by assuming the solution y = e^rt not equla to Zero.

So y’ = r e^rt

      y’’ = r²e^rt

so substitute in (1)

r² e^rt - 2 r e^rt + e^rt = 0

e^rt (r² – 2r + 1) = 0

e^rt is not equal to zero so r² – 2r + 1 = 0

r² – 2r + 1 = 0

r² – r – r + 1 = 0

r(r – 1) -1 (r – 1) = 0

(r – 1) (r – 1) = 0

r₁ = r₂ = 1

so when the characteristic equation roots are real and repeat then homogeneous solution is in the form as given below.

• Yc = C­₁e^r1t + C₂te^r2t
• Yc = C_­1e^t + C₂te^t

So now we need to find the particular solution

We have a non homogeneous term g(t) = cos(2t)

Now the differential cycle of g’(t) is = -2 sin(2t)

And g’’(t) = -4 cos(2t)

so we can assume Yp = A sin(2t) + B cos(2t)

then Y’p = 2A cos(2t) – 2B sin(2t)

and Y’’p = -4Asin(2t) – 4Bcos(2t)

Now substitute in given IVP

Yp’’ - 2Y’p + Yp = cos(2t)

-4Asin(2t) – 4Bcos(2t) – 2[2A cos(2t) – 2B sin(2t)] + A sin(2t) + B cos(2t) = cos(2t)

-4Asin(2t) – 4Bcos(2t) –4A cos(2t) + 4B sin(2t) + A sin(2t) + B cos(2t) = cos(2t)

-4Asin(2t) – 4Bcos(2t) –4A cos(2t) + 4B sin(2t) + A sin(2t) + B cos(2t) = cos(2t)

Sin(2t) (-3A + 4B) + cos(2t) (-4A – 3B) = cos(2t)

Now equal the cos(2t) coefficients then -4A – 3B = 1 --------(1)

Now equal the sin(2t) coefficients then -3A + 4B = 0 --------(2)

Now do 3 x (1) - 4 x (2)   then -12A – 9B +12A – 16B = 3 + 0

-25B = 3 then B = -3/25

Now from (1)    -4A – 3(-3/25) = 1

-4A + (9/25) = 1

-4A = 1 – (9/25)

-4A = 16/25

Then A = -4/25

So particular solution Yp = (-4/25) sin(2t) - (3/25) cos(2t)

And the most general solution Y = Yc + Yp

• Y = C_­1e^t + C₂te^t - (4/25) sin(2t) - (3/25) cos(2t)

Now we have initial conditions y(0) = 0 and y’(0) =0

• Y(t) = C_­1e^t + C₂te^t - (4/25) sin(2t) - (3/25) cos(2t)
• Y(0) = C₁e⁰ + C­₂(0)e⁰ - (4/25) sin(0) - (3/25) cos(0)
• 0 = C₁ - (4/25) (0) - (3/25) (1)
• 0 = C₁ - (3/25)
• C₁ = 3/25

• Now Y’(t) = C_­1e^t + C₂e^t + C₂te^t - (8/25) sin(2t) + (6/25) cos(2t)

• So Y’(0) = C_­1e⁰ + C₂e⁰ + C₂(0)e⁰ - (8/25) sin(0) + (6/25) cos(0)

0 = C_­1 + C₂ - (8/25) (0) + (6/25) (1)

0 = C_­1 + C₂ + (6/25)

C_­1 + C₂ = -(6/25)

(3/25) + C₂ = -6/25

C₂ = (-6/25) – (3/25)

C₂ = (-9/25)

• Y(t) = (3/25)e^t - (9/25)te^t - (4/25) sin(2t) - (3/25) cos(2t)