answer:To solve the given nonlinear differential equation, we can use the method of separation of variables. First, rewrite the equation as: frac{dy}{y^2} = dx(x^2) Now, integrate both sides: int frac{dy}{y^2} = int dx(x^2) The left side is a simple integration: -frac{1}{y} = int x^2 dx Now, integrate the right side: -frac{1}{y} = frac{1}{3}x^3 + C Now, we need to find the constant of integration, C. We know that y(0) = 1. Plug in these values: -frac{1}{1} = frac{1}{3}(0)^3 + C -1 = C So, the equation becomes: -frac{1}{y} = frac{1}{3}x^3 - 1 Now, we need to find the value of y when x = 1. Plug in x = 1: -frac{1}{y} = frac{1}{3}(1)^3 - 1 -frac{1}{y} = frac{1}{3} - 1 -frac{1}{y} = -frac{2}{3} Now, solve for y: y = frac{3}{2} So, the value of y when x = 1 is frac{3}{2}.

answer:To solve the given first-order nonlinear differential equation, we can use the method of separation of variables. The differential equation is: dy/dt = y(5-2y) Rearrange the equation to separate the variables: dy/(y(5-2y)) = dt Now, integrate both sides of the equation: ∫(1/(y(5-2y))) dy = ∫ dt To solve the integral on the left side, we can use partial fraction decomposition. Let: 1/(y(5-2y)) = A/y + B/(5-2y) Multiplying both sides by y(5-2y) to clear the denominators, we get: 1 = A(5-2y) + B(y) Now, we can solve for A and B by choosing appropriate values for y: 1. Let y = 0: 1 = 5A => A = 1/5 2. Let y = 5/2: 1 = B(5/2) => B = 2/5 Now we can rewrite the integral as: ∫(1/5(1/y) + 2/5(1/(5-2y))) dy = ∫ dt Integrate both sides: (1/5)∫(1/y) dy + (2/5)∫(1/(5-2y)) dy = ∫ dt (1/5)ln|y| - (1/5)ln|5-2y| = t + C Now, we can use the initial condition y(0) = 1: (1/5)ln|1| - (1/5)ln|5-2(1)| = 0 + C 0 - (1/5)ln|3| = C C = -(1/5)ln(3) Now we have the general solution: (1/5)ln|y| - (1/5)ln|5-2y| = t - (1/5)ln(3) To find the behavior of the system as t increases to infinity, we can analyze the equation: As t -> ∞, the left side of the equation must also increase. Since ln|y| and ln|5-2y| are both logarithmic functions, they will increase slowly as their arguments increase. However, ln|5-2y| has a negative coefficient, so it will decrease as its argument increases. This means that as t -> ∞, y must approach the value that makes the term ln|5-2y| as small as possible, which occurs when y -> 5/2. Therefore, as t increases to infinity, the behavior of the system is such that y(t) approaches 5/2.

answer:To find the position of the particle at time t = 5 seconds, we first need to find the acceleration, then the velocity, and finally the position. 1. Acceleration: We can find the acceleration using Newton's second law, F = ma. Since F(x) = -kx^3, we have: ma = -kx^3 a = -frac{k}{m}x^3 Since the mass m = 2 kg, we have: a = -frac{k}{2}x^3 2. Velocity: To find the velocity, we need to integrate the acceleration with respect to time: v(t) = int a dt = int -frac{k}{2}x^3 dt However, we cannot directly integrate this expression, as it involves both x and t. To proceed, we need to express the acceleration as a function of time. We can do this by using the chain rule: frac{dv}{dt} = frac{dv}{dx} cdot frac{dx}{dt} Since frac{dx}{dt} = v, we have: frac{dv}{dt} = -frac{k}{2}x^3 cdot v Now, we can separate variables and integrate: int frac{dv}{v} = -frac{k}{2} int x^3 dt We don't have enough information to proceed further, as the constant k is unknown. The problem statement is incomplete, and we cannot find the position of the particle at time t = 5 seconds without more information.

answer:To solve the given second-order nonlinear differential equation using the fourth-order Runge-Kutta method, we first need to convert it into a system of first-order differential equations. Let's introduce two new variables: v = y' w = y'' Now, we have the following system of first-order differential equations: y' = v v' = w w' = sin(x) - v^2 - y*sin(y) with the initial conditions y(0) = 1, v(0) = 0. Now, we can apply the fourth-order Runge-Kutta method to solve this system of equations. Here's a Python code using the `scipy` library to solve the problem and plot the solution: ```python import numpy as np import matplotlib.pyplot as plt from scipy.integrate import solve_ivp def system_of_equations(t, y): y, v, w = y dydt = v dvdt = w dwdt = np.sin(t) - v**2 - y*np.sin(y) return [dydt, dvdt, dwdt] # Initial conditions y0 = [1, 0, 0] # Time interval t_span = (0, 10) t_eval = np.linspace(0, 10, 1000) # Solve the system of equations using the fourth-order Runge-Kutta method sol = solve_ivp(system_of_equations, t_span, y0, method='RK45', t_eval=t_eval) # Plot the solution plt.plot(sol.t, sol.y[0], label='y(x)') plt.xlabel('x') plt.ylabel('y(x)') plt.title('Solution of the nonlinear differential equation') plt.legend() plt.show() ``` This code will solve the given differential equation in the interval [0, 10] and plot the solution y(x). Note that the exact solution cannot be expressed in a closed form, so the numerical solution is the best we can do.