Use the method of "undetermined coefficients" to solve the 2nd ODE #y''+4y'+4y=e^(-2x)sin2x# ?

Answer 1

# y(x) = Axe^(-2x)+Be^(-2x) -1/4e^(-2x)sin2x #

We have:

# y''+4y'+4y = e^(-2x)sin2x #
This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, #y_p# of the non-homogeneous equation.

Complementary Role

The equation homogeneous linked to [A] is

# y''+4y'+4y #

Additionally, the related auxiliary equation is:

# m^2+4m+4 = 0 # # (m+2)^2 = 0 #
Which has a repeated real root #m=-2#

Consequently, the homogeneous equation's solution is:

# y_c = (Ax+B)e^(-2x) #

Specific Resolution

For this specific equation [A], the following is a likely solution:

# y = e^(-2x)(acos2x + bsin2x) #
Where #a,b# are constant coefficients to be determined. Let us assume the above solution works, in which case be differentiating wrt #x# we have:
# y' \ \ = e^(-2x)(-2asin2x + 2bcos2x) -2e^(-2x)(acos2x + bsin2x)# # \ \ \ \ \ = e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x)#

And:

# y'' = e^(-2x)(-2(2b-2a)sin2x - 2(2a+2b)cos2x) -2 e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x) #
# \ \ \ \ \ = e^(-2x)(-8b)cos2x + (8a)sin2x) #
Substituting into the initial Differential Equation #[A]# we get:
# {e^(-2x)(-8b)cos2x + (8a)sin2x)} + 4{e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x)} + 4{e^(-2x)(acos2x + bsin2x)} = e^(-2x)sin2x #
Equating coefficients of #cos2x# and #sin2x# we get:
#cos2x: -8b + 8b-8a + 4a =0 # #sin2x: 8a - 8a-8b+4b=1 #

When we solve concurrently, we have:

# a = 0, b=-1/4#

Thus, we arrive at the specific solution:

# y_p = -1/4e^(-2x)sin2x #

Overall Resolution

which ultimately results in the GS of [A}

# y(x) = y_c + y_p # # \ \ \ \ \ \ \ = (Ax+B)e^(-2x) -1/4e^(-2x)sin2x # # \ \ \ \ \ \ \ = Axe^(-2x)+Be^(-2x) -1/4e^(-2x)sin2x #
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Answer 2

To solve the second-order ordinary differential equation (ODE) (y'' + 4y' + 4y = e^{-2x}\sin(2x)) using the method of undetermined coefficients, follow these steps:

  1. Find the complementary solution by solving the homogeneous equation (y'' + 4y' + 4y = 0). The characteristic equation is (r^2 + 4r + 4 = 0), which has a repeated root at (r = -2). So, the complementary solution is (y_c(x) = (C_1 + C_2x)e^{-2x}).

  2. Next, consider the particular solution (y_p(x)) for the non-homogeneous part (e^{-2x}\sin(2x)). Since (e^{-2x}\sin(2x)) is a product of an exponential function and a trigonometric function, we assume (y_p(x) = Ae^{-2x}\sin(2x) + Be^{-2x}\cos(2x)), where (A) and (B) are undetermined coefficients.

  3. Differentiate (y_p(x)) twice to find (y_p'') and (y_p').

  4. Substitute (y_p), (y_p'), and (y_p'') into the original ODE (y'' + 4y' + 4y = e^{-2x}\sin(2x)).

  5. Equate the coefficients of like terms on both sides of the equation.

  6. Solve the resulting system of equations to find the values of (A) and (B).

  7. Combine the complementary solution (y_c(x)) and the particular solution (y_p(x)) to get the general solution (y(x) = y_c(x) + y_p(x)).

  8. Finally, if initial conditions are given, use them to determine the values of the constants (C_1) and (C_2).

This completes the solution process.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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