IDZ 11.2 – Option 5. Solutions Ryabushko A.P.

1. First, let's find a particular solution to the differential equation y´´= 4cos2x. Integrating twice, we get y = -cos2x + Ax + B, where A and B are arbitrary constants. Next, substituting the initial conditions y(0) = 1 and y´(0) = 3, we find the values ​​of the constants A and B: A = (3+cos0.5π)/0.5π ≈ 6.45 and B = 1 + cos0.5π - 6.45*0.5π ≈ 0.22. Thus, the partial solution has the form y = -cos2x + 6.45x + 0.22. Substituting the value x = π/4, we get y(π/4) ≈ 4.12.

2. Consider the differential equation y´´xlnx = y´. Let us make the replacement y´ = v, then y´´ = v´ + v/x. Substituting this into the original equation, we get v´ + v/x = v, which is equivalent to v´ = -v/x. Let's solve this equation using the method of separation of variables: v´/v = -1/x dx, ln|v| = -ln|x| + C1, where C1 is an arbitrary constant. Thus, v = C/x, where C is an arbitrary constant. Returning to the original variables, we obtain y´ = C/x, y = C ln|x| + D, where D is an arbitrary constant. In total, the general solution of the differential equation has the form y = C ln|x| + D.

3. Consider the differential equation y´´tgy = 2y´2. Let us make the replacement y´ = v, then y´´ = v´´tg(x) + (v´)2sec2(x). Substituting this into the original equation, we get v´´tg(x) + (v´)2sec2(x) = 2v2. To simplify the equation, we use the substitution u = v2, then u´ = 2vv´. Substituting this into the original equation, we get u´/2 = utg(x) + usec2(x), which is equivalent to u´/2u = tan(x) + sec2(x). Let's solve this equation using the method of separation of variables: ln|u|/2 = ln|sin(x)| - ln|cos(x)| + C1, where C1 is an arbitrary constant. Thus, u = C sin(x)/cos2(x), where C is an arbitrary constant. Returning to the original variables, we obtain v = ±√(C sin(x)/cos2(x)), y = ∫v dx = ±∫√(C sin(x)/cos2(x)) dx. To integrate, we use the substitution t = cos(x), then y = ±∫√(C/t) dt = ±2√Ct + C1, where C1 is an arbitrary constant. In total, the general solution of the differential equation has the form y = ±2√Ccos(x) + C1.

4. Consider the equation (x/√x2-y2-1)dx-ydy/√x2-y2=0. To solve, we use the replacement y = xz, then y´ = z + xz´. Substituting this into the original equation, we get (x/√x^2 - x^2z^2 - 1)dx - (xz + x^2z´)/√(x^2 - x^2z^2) dz = 0. Taking x out of the first term and combining the fractions of the second term, we get (1/√(1 - z^2))dz = (1/x)(√(x^2 - x^2z^2))dx. Let's solve this equation using the separation of variables method: ∫(1/√(1 - z^2))dz = ∫(1/x)(√(x^2 - x^2z^2))dx, arcsin(z) = ln |x| + C1, where C1 is an arbitrary constant. Thus, z = sin(ln|x| + C1), y = xz = xsin(ln|x| + C1). In total, the solution to the equation is the curve y = x*sin(ln|x| + C1).

5. Let us write the equation of a curve passing through point A(-2, 1), if the angular coefficient of the tangent at any point is equal to the ordinate of this point, increased by 5 times. The slope of the tangent is equal to the derivative of the function at a given point. Let y = f(x) be the equation of the desired curve. Then the condition on the slope can be written as f´(x) = 5f(x). Let's solve this equation using the method of separation of variables: f´(x)/f(x) = 5 dx, ln|f(x)| = 5x + C1, where C1 is an arbitrary constant. Substituting the coordinates of point A(-2, 1), we find C1 = ln|1/2|. Thus, the equation of the desired curve has the form y = f(x) = Ce^(5x), where C = 1/2. In total, the equation of the curve passing through the point A(-2, 1) and satisfying the given condition has the form y = (1/2)e^(5x).

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IDZ 11.2 – Option 5. Solutions Ryabushko A.P.

This product is a digital product in HTML format, which contains solutions to tasks of option 5 of Individual Homework No. 11.2 on mathematical analysis, developed by the author Ryabushko A.P.

The product is intended for students studying mathematical analysis, as well as for teachers using this educational material.

Solutions to assignments are presented in a beautifully designed HTML document, which allows you to conveniently and quickly familiarize yourself with the content and go to the desired section.

By purchasing this product, you are getting a quality product that will help you better understand calculus and complete assignments successfully.

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IDZ 11.2 – Option 5. Solutions Ryabushko A.P. is a collection of solutions to problems in mathematical analysis. This version contains problems on differential equations and curves on the plane.

1. In the first problem, you need to find a particular solution to the differential equation and calculate the value of the function y=φ(x) at x=x0 accurate to two decimal places. The equation has the form: 1.5 y´´= 4cos2x, x0 = π/4, y(0) = 1, y´(0) = 3. The solution to this problem is presented in the collection and designed using the formula editor Microsoft Word 2003.

2. The second problem requires finding a general solution to a differential equation that can be reduced in order. The equation looks like: 2.5 y´´xlnx = y´. There is no solution to this problem in this version.

3. The third problem requires solving the Cauchy problem for a differential equation that can be reduced in order. The equation has the form: 3.5 y´´tgy = 2y´2, y(1) = π/2, y´(1) = 2. The solution to this problem is presented in the collection and designed using the formula editor Microsoft Word 2003.

4. The fourth problem requires integrating this equation: 4.5 (x/√x2-y2-1)dx-ydy/√x2-y2=0. There is no solution to this problem in this version.

5. In the fifth problem, it is required to write the equation of a curve passing through the point A(x0, y0), if it is known that the slope of the tangent at any point is equal to the ordinate of this point, increased by k times. Point A has coordinates A(−2, 1), k = 5. The solution to this problem is not available in this version.

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