1. To solve this problem, let's find a general solution to the differential equation y´´´= cos2x. Integrating this equation three times, we obtain y´´= (1/2)sin2x + C1, y´= -(1/4)cos2x + C1x + C2, where C1 and C2 are arbitrary constants. Integrating the last expression, we get y= (1/8)sin2x - (1/4)xsin2x - (1/8)cos2x + C1x2 + C2x + C3, where C3 is another arbitrary constant. Substituting the initial conditions, we obtain a system of equations: C1 = 0, C2 = 1/8, C3 = 23/64. Thus, the partial solution has the form y= (1/8)sin2x - (1/4)xsin2x - (1/8)cos2x + (23/64).

2. To find the general solution to this differential equation y´´ = −x/y´, multiply both sides by y´ and integrate twice: y´dy´´ = -xdy´, y´´dy´ = -xdy, y´´dy ´ - y´dy´´ = xdy. When integrating by parts, we obtain y´´y´ - (y´)2/2 = -(x2/2) + C1, where C1 is an arbitrary constant. Solving this equation for y´, we obtain y´ = ±sqrt(C1 - x2) and, accordingly, y = ±(1/2)int(sqrt(C1 - x2)dx) + C2, where C2 is another arbitrary constant . Thus, the general solution is y = ±(1/2)(sin^(-1)x + xsqrt(1 - x2)) + C2.

3. To solve the Cauchy problem for the differential equation y´´ = 1 − y´2 with the initial conditions y(0) = 0, y´(0) = 0, we make the substitution y´ = p(y) and obtain the equation p´dp/ dy = 1 - p2. Integrating this equation, we obtain p = sin(y + C1), where C1 is an arbitrary constant. Substituting this expression into the equation y´ = p(y), we obtain y´ = sin(y + C1). Integrating this equation, we get y = -cos(y + C1) + C2, where C2 is another arbitrary constant. Substituting the initial conditions, we obtain C1 = 0, C2 = 1. Thus, the solution to the Cauchy problem has the form y = 1 - cos(y).

4. This differential equation can be reduced to the form dx/dy = -(x2 + 2y2)/(xy(x + 2y)), which can be solved by the separation of variables method. After simple algebraic transformations, we obtain: (y + x)dx + 2ydy = 0. Integrating this equation, we obtain x2 + 2y2 + 2xy = C, where C is the integration constant.

5. Let the equation of the curve passing through the point A(x0, y0) have the form y = kx + b, where k and b are coefficients. Then the angular coefficient of the tangent at the point (x0, y0) is equal to k, and the angular coefficient of the straight line connecting point A with the origin is equal to y0/x0. From the conditions of the problem it follows that k = n*(y0/x0), where n is a given number. Thus, the equation of the curve is y = n*(y0/x0)x + b. Substituting the coordinates of point A, we get y = 9(-4/6)*x + 10, that is, the equation of the curve is y = (-6/2)x + 10, which is equivalent to y = -3x + 10.

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IDZ 11.2 – Option 9. Solutions of Ryabushko A.P.

1. Find a particular solution to the differential equation

The differential equation y´´´= cos2x is given. Let's find a general solution by integrating this equation three times:

y´´= (1/2)sin2x + C1,

y´= -(1/4)cos2x + C1x + C2,

y= (1/8)sin2x - (1/4)xsin2x - (1/8)cos2x + C1x2 + C2x + C3,

where C1, C2 and C3 are arbitrary constants.

Substituting the initial conditions y(0) = 1, y´(0) = −1/8, y´´(0) = 0, we obtain C1 = 0, C2 = 1/8, C3 = 23/64.

Thus, a particular solution has the form:

y= (1/8)sin2x - (1/4)xsin2x - (1/8)cos2x + (23/64)

Value of the function at x=π:

y(π) = (1/8)sin2π - (1/4)πsin2π - (1/8)cos2π + (23/64) ≈ 0.17

2. Find the general solution to the differential equation

The differential equation y´´ = −x/y´ is given. To find the general solution, multiply both sides by y´ and integrate twice:

y´dy´´ = -xdy´,

y´´dy´ = -xdy,

y´´dy´ - y´dy´´ = xdy.

When integrating by parts we get:

y´´y´ - (y´)2/2 = -(x2/2) + C1,

where C1 is an arbitrary constant.

Solving this equation for y´, we get:

y´ = ±sqrt(C1 - x2)

and, accordingly,

y = ±(1/2)int(sqrt(C1 - x2)dx) + C2,

where C2 is another arbitrary constant.

Thus, the general solution looks like:

y = ±(1/2)(sin^(-1)x + xsqrt(1 - x2)) + C2.

3. Solve the Cauchy problem for a differential equation

Given the differential equation y´´ = 1 − y´2 and the initial conditions y(0) = 0, y´(0) =0. Let's make the replacement y´ = p(y) and get the equation:

p´dp/dy = 1 - p2.

Integrating this equation, we get:

p = sin(y + C1),

where C1 is an arbitrary constant.

Substituting this expression into the equation y´ = p(y), we get:

y´ = sin(y + C1).

Integrating this equation, we get:

y = -cos(y + C1) + C2,

where C2 is another arbitrary constant.

Using the initial conditions y(0) = 0, y´(0) = 0, we obtain:

C1 = 0, C2 = 1.

Thus, the solution to the Cauchy problem has the form:

y = -cos(y) + 1,

or, equivalently,

cos(y) = 1 - y.

Thus, the solution to the Cauchy problem is given by the equation cos(y) = 1 - y, where y = y(x) is the solution to the differential equation satisfying the initial conditions y(0) = 0, y´(0) = 0.

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IDZ 11.2 – Option 9. Solutions Ryabushko A.P. is a set of solutions to problems in mathematical analysis and differential equations, completed by the author Ryabushko A.P.

This project contains solutions to the following problems:

1. Find a particular solution to the differential equation and calculate the value of the resulting function y=φ(x) at x=x0 accurate to two decimal places. The differential equation has the form y´´´= cos2x, the initial conditions are given y(0) = 1, y´(0) = −1/8, y´´(0) = 0, it is required to find a solution at x = π.

2. Find a general solution to a differential equation that can be reduced in order. The equation is y´´ = −x/y´.

3. Solve the Cauchy problem for a differential equation that admits a reduction in order. The equation has the form y´´ = 1 − y´2, the initial conditions y(0) = 0, y´(0) = 0 are given.

4. Integrate the equation x(2x2 + y2) + y(x2 + 2y2)y’= 0.

5. Write down the equation of a curve passing through the point A(x0, y0), if it is known that the angular coefficient of the tangent at any point is n times greater than the angular coefficient of the straight line connecting the same point with the origin. Given a point A(−6, 4) and n = 9.

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

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