IDZ 11.3 – Option 13. Solutions Ryabushko A.P.

  1. Let's find the general solution to the differential equations: a) 9y΄΄+ 6y΄ + y = 0; Characteristic equation: 9λ² + 6λ + 1 = 0 D = 36 - 4*9 = 0 λ₁ = λ₂ = -1/3 y₁(x) = e^(-x/3) y₂(x) = xe^(-x /3) Y(x) = c₁e^(-x/3) + c₂xe^(-x/3)
  2. b) y΄΄− 4y΄− 21y = 0; Characteristic equation: λ² - 4λ - 21 = 0 D = 16 + 84 = 100 λ₁ = 7, λ₂ = -3 y₁(x) = e^(7x) y₂(x) = e^(-3x) Y(x) = c₁e^(7x) + c₂e^(-3x)

    c) y΄΄+ y = 0 Characteristic equation: λ² + 1 = 0 D = -4 λ₁ = i, λ₂ = -i y₁(x) = cos(x) y₂(x) = sin(x) Y(x ) = c₁cos(x) + c₂sin(x)

    1. Let's find the general solution of the differential equation: y΄΄ – 8y΄ + 12y = 36x^4 – 96x^3 + 24x^2 + 16x – 2 Characteristic equation: λ² - 8λ + 12 = 0 D = 64 - 48 = 16 λ₁ = 6 , λ₂ = 2 y₁(x) = e^(6x) y₂(x) = e^(2x) Yh(x) = c₁e^(6x) + c₂e^(2x)

    Let us find a particular solution y_p(x) of the inhomogeneous equation by the method of indefinite coefficients: y_p(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E y΄(x) = 4Ax^3 + 3Bx^2 + 2Cx + D y΄΄(x) = 12Ax^2 + 6Bx + 2C Substitute into the equation: 12Ax^2 + 6Bx + 2C - 8(4Ax^3 + 3Bx^2 + 2Cx + D) + 12(Ax^4 + Bx^3 + Cx^2 + Dx + E) = 36x^4 – 96x^3 + 24x^2 + 16x – 2 12A = 36, -32A + 6B = -96, 20A - 24B + 2C = 24, -8A + 12B - 8C + 2D = 16, 12A - 8B + 12C - 8D + 12E = -2 A = 3, B = 9/4, C = -27/8, D = -21/16, E = -131/288 Yp( x) = 3x^4 + (9/4)x^3 - (27/8)x^2 - (21/16)x - 131/288 Y(x) = Yh(x) + Yp(x)

    1. Let's find the general solution of the differential equation: y΄΄ + 4y΄ + 4y = 6e^(-2x) Characteristic equation: λ² + 4λ + 4 = 0 D = 16 - 16 = 0 λ₁ = λ₂ = -2 y₁(x) = e ^(-2x) y₂(x) = xe^(-2x) Yh(x) = c₁e^(-2x) + c₂xe^(-2x)

    Let us find a particular solution y_p(x) of the inhomogeneous equation by the method of variation of constants: Let us present the solution in the form y(x) = u(x)e^(-2x) y΄(x) = u΄(x)e^(-2x) - 2u(x)e^(-2x) y΄΄(x) = u΄΄(x)e^(-2x) - 4u΄(x)e^(-2x) + 4u(x)e^(- 2x) Substitute into the equation: u΄΄(x)e^(-2x) + 2u΄(x)e^(-2x) = 6 Integrate both sides: u΄(x)e^(-2x) = 3x + c₁ u(x) = -3/2x^2 - c₁/2x + c₂ Yp(x) = (-3/2x^2 - c₁/2x + c₂)e^(-2x) Y(x) = Yh( x) + Yp(x)

    1. Let us find a particular solution to the differential equation that satisfies the given initial conditions: y΄΄− 6y΄ + 25y = (32x – 12)sin(3x) – 36xcos(3x), y(0) = 4, y΄(0) = 0 Characteristic equation: λ² - 6λ + 25 = 0 D = 36 - 100 = -64 λ₁ = 3 - 4i, λ₂ = 3 + 4i y₁(x) = e^(3x)sin(4x) y₂(x) = e^( 3x)cos(4x) Yh(x) = e^(3x)(c₁sin(4x) + c₂cos(4x))

    Let us find a particular solution y_p(x) of the inhomogeneous equation by the method of indefinite coefficients: y_p(x) = A(x)sin(3x) + B(x)cos(3x) y΄(x) = 3A(x)cos(3x) - 3B(x)sin(3x) + 32 y΄΄(x) = -9A(x)sin(3x) - 9B(x)cos(3x) Substitute into the equation: -9A(x)sin(3x) - 9B (x)cos(3x) - 6(3A(x)cos(3x) - 3B(x)sin(3x) + 32) + 25(A(x)sin(3x) + B(x)cos(3x) ) = (32x - 12)sin(3x) - 36xcos(3x) -18A(x) + 25A(x) = -36x, -18B(x) + 25B(x) = 32 A(x) = 4x/7 , B(x) = 32/7 Yp(x) = (4x/7)sin(3x) + (32/7)cos(3x) Y(x) = Yh(x) + Yp(x) Substitute the initial conditions and find the constants: Y(0) = c₂ = 4 Y΄(0) = 3c₁ + 32/7 = 0 => c₁ = -32/21 Y(x) = e^(3x)(-4/3sin(4x ) + 32/21cos(4x)) + (4x/7)sin(3x) + (32/7)cos(3x)

    1. Let us define and write the structure of a particular solution y*(x) of a linear inhomogeneous differential equation by the form of the function f(x): y΄΄– 6y΄ + 9y = f(x) a) f(x) = (x – 2)e^ (3x) Characteristic equation: λ² - 6λ + 9 = 0 (λ - 3)² = 0 => λ₁ = λ₂ = 3 y₁(x) =

    This product is a digital product offered in the Digital Product Store. This is a solution to problem IDZ 11.3, Option 13, prepared by the author Ryabushko A.P.

    Solution IDZ 11.3 - Option 13 includes a detailed description of the solution to the problem, as well as a step-by-step explanation of all the steps required to solve it.

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    This product is a solution to the problem IDZ 11.3 – Option 13, which includes a detailed description of the solution to differential equations and methods for solving them. The solution was prepared by the author Ryabushko A.P. and is designed in a beautiful html format, which makes it easy to read and understand.

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IDZ 11.3 – Option 13. Solutions Ryabushko A.P. is a set of solutions to differential equations consisting of five problems. Each problem is a differential equation that needs to be solved.

In the first problem, you need to find a general solution to the differential equation, which has the form 9y΄΄+ 6y΄ + y = 0 in point a), the form y΄΄− 4y΄− 21y = 0 in point b) and the form y΄΄+ y = 0 in point c).

In the second problem, it is necessary to find a general solution to the differential equation, which has the form y΄΄ – 8y΄ + 12y = 36x4 – 96x3 + 24x2 + 16x – 2.

The third problem requires finding a general solution to the differential equation, which has the form y΄΄ + 4y΄ + 4y = 6e–2x.

In the fourth problem, it is necessary to find a particular solution to the differential equation, which has the form y΄΄− 6y΄ + 25y = (32x – 12)sin3x – 36xcos3x, provided that y(0) = 4 and y΄(0) = 0.

In the fifth problem, it is required to determine and write down the structure of a particular solution y* of a linear inhomogeneous differential equation in terms of the form of the function f(x), which has the form y΄΄– 6y΄ + 9y = f(x). In point a) the function f(x) is equal to (x – 2)e3x, and in point b) the function f(x) is equal to 4cosx.

Each problem has a detailed solution, designed in Microsoft Word 2003 using the formula editor.


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