Question

Y'''+2y''-y'=3x^2+14x+11

Answer 1

Посмотрите предложенное решение; оформление не соблюдалось.

Answer 2

To solve the given third-order linear homogeneous differential equation:

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y''' + 2y'' - y' = 3x^2 + 14x + 11

We first need to find the complementary solution by solving the associated homogeneous equation:

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y''' + 2y'' - y' = 0

Once we have the complementary solution, we can find a particular solution to the original non-homogeneous equation using the method of undetermined coefficients or variation of parameters.

Step 1: Find the Complementary Solution

The characteristic equation for the homogeneous equation is:

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r^3 + 2r^2 - r = 0

Now, we solve the characteristic equation:

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r(r^2 + 2r - 1) = 0

Using either the quadratic formula or factoring, we can find the roots of the characteristic equation:

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r1 = 0
r2 = (-2 + √(2^2 - 4*(-1))) / 2
r2 = (-2 + √(4 + 4)) / 2
r2 = (-2 + √8) / 2
r2 = (-2 + 2√2) / 2
r2 = -1 + √2

r3 = (-2 - √(2^2 - 4*(-1))) / 2
r3 = (-2 - √(4 + 4)) / 2
r3 = (-2 - √8) / 2
r3 = (-2 - 2√2) / 2
r3 = -1 - √2

The complementary solution is a linear combination of the form:

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y_c = C1 * e^(r1 * x) + C2 * e^(r2 * x) + C3 * e^(r3 * x)

where C1, C2, and C3 are constants to be determined based on initial or boundary conditions.

Step 2: Find a Particular Solution

For the particular solution, we'll assume it has a form similar to the right-hand side of the non-homogeneous equation:

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y_p = Ax^2 + Bx + C

where A, B, and C are constants to be determined.

Now, we find the derivatives of y_p:

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y_p' = 2Ax + B
y_p'' = 2A
y_p''' = 0

Substitute the derivatives of y_p into the non-homogeneous equation:

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0 + 2(2A) - (2Ax + B) = 3x^2 + 14x + 11

Simplify:

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4A - 2Ax - B = 3x^2 + 14x + 11

Now, equate the coefficients of like powers of x on both sides:

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-2A = 3    => A = -3/2
4A - B = 14 => B = 4A - 14 = 4 * (-3/2) - 14 = -6 - 14 = -20

Therefore, the particular solution is:

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y_p = (-3/2)x^2 - 20x + C

Step 3: General Solution

Now, combine the complementary solution and the particular solution to obtain the general solution:

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y = y_c + y_p

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y = C1 * e^(0) + C2 * e^((-1 + √2) * x) + C3 * e^((-1 - √2) * x) - (3/2)x^2 - 20x + C

Simplifying:

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y = C1 + C2 * e^((-1 + √2) * x) + C3 * e^((-1 - √2) * x) - (3/2)x^2 - 20x + C

Where C1, C2, C3, and C are constants that can be determined using initial or boundary conditions, if provided.