#y'' + 9y = 2x^2 - 5#, How about solution. (#y_h#=?, #y_p=#?)

Answer 1

# y(x) = Acos(3x)+Bsin(3x) + 2/9x^2 -49/91 #

We are looking for a fix for

# y'' + 9y = 2x^2 - 5 # ..... [A]
This is a second order linear non-Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the polynomial 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'' + 9y = 0 # ..... [B]

Additionally, the related auxiliary equation is:

# m^2+9 = 0#

Consequently, we have the answers:

# m = +--1 \ \ \ \ # (pure imaginary)

Parts of the solution are determined by the auxiliary equation's roots; if these parts are linearly independent, the solutions' superposition forms the complete general solution.

Thus, the homogeneous equation [B] has the following solution:

# y = e^(0x)(Acos(3x)+Bsin(3x) ) # # \ \ = Acos(3x)+Bsin(3x) #

Specific Resolution

To identify a specific non-homogeneous equation solution:

# y'' + 9y = f(x) \ \ # with #f(x) = 2x^2 - 5 #

Thus, it is probably best if we search for a solution of this kind:

# y = ax^2+bx+c # ..... [C]
Where the constants #a,b,c,d# is to be determined by direct substitution and comparison:
Differentiating [C] wrt #x# twice we get:
# y^((1)) = 2ax+b # # y^((2)) = 2a #

When we enter these outcomes into the DE [A], we obtain:

# (2a) + 9(ax^2+bx+c) = 2x^2 - 5 #

Calculating coefficients gives us:

# x^2: 9a = 2 => a = 2/9 # # x^1: 9b=0=> b =0 # # x^0: 2a+9c = -5 => c = -49/91 #

Thus, we arrive at the specific solution:

# y_p = 2/9x^2 -49/91 #

Overall Resolution

which ultimately results in the GS of [A}

# y(x) = y_c + y_p # # \ \ \ \ \ \ \ = Acos(3x)+Bsin(3x) + 2/9x^2 -49/91 #
Note this solution has #2# constants of integration and #2# linearly independent solutions, hence by the Existence and Uniqueness Theorem their superposition is the General Solution
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Answer 2

The homogeneous solution (y_h) is y_h = C₁cos(3x) + C₂sin(3x), where C₁ and C₂ are arbitrary constants.

The particular solution (y_p) is y_p = (1/18)*x^2 - (5/18).

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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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