answer:To solve the given differential equation dy/dx = 2x, we need to integrate both sides with respect to x. ∫(dy/dx) dx = ∫2x dx Integrating both sides, we get: y(x) = x^2 + C Now, we are given the initial condition y(0) = 4. We can use this to find the constant C. y(0) = (0)^2 + C = 4 C = 4 So, the particular solution to the differential equation is: y(x) = x^2 + 4 Now, we need to find the value of y when x = 3. y(3) = (3)^2 + 4 y(3) = 9 + 4 y(3) = 13 Therefore, the value of y when x = 3 is 13.

answer:To find the particular solution, we need to integrate the given differential equation and apply the initial condition. 1. Integrate the differential equation: dy/dx = 4x - 3 Integrating both sides with respect to x, we get: ∫dy = ∫(4x - 3)dx y(x) = 2x^2 - 3x + C 2. Apply the initial condition y(0) = 5: y(0) = 2(0)^2 - 3(0) + C = 5 C = 5 3. Write the particular solution: y(x) = 2x^2 - 3x + 5

answer:Let S(t) be the amount of salt in the tank at time t (in minutes). The rate of salt flowing into the tank is 2 gal/min * 0.05 lb/gal = 0.1 lb/min. The rate of salt flowing out of the tank is proportional to the concentration of salt in the tank, which is S(t)/100 lb/gal, and the flow rate of 2 gal/min. So, the rate of salt flowing out is (S(t)/100) * 2 lb/min. Now, we can write a differential equation for the amount of salt in the tank as a function of time: dS/dt = rate of salt flowing in - rate of salt flowing out dS/dt = 0.1 - (S(t)/100) * 2 To solve this differential equation, we can use an integrating factor. The integrating factor is e^(∫-2/100 dt) = e^(-t/50). Multiply both sides of the differential equation by the integrating factor: e^(-t/50) dS/dt - (2/100)e^(-t/50) S(t) = 0.1e^(-t/50) Now, the left side of the equation is the derivative of the product S(t)e^(-t/50). So, we can integrate both sides with respect to t: ∫(d/dt[S(t)e^(-t/50)]) dt = ∫0.1e^(-t/50) dt Integrating both sides, we get: S(t)e^(-t/50) = -5e^(-t/50) + C Now, we can solve for the initial condition, S(0) = 1 lb: 1 = -5 + C C = 6 So, the equation for S(t) is: S(t) = -5e^(-t/50) + 6e^(t/50) Now, we can find the amount of salt in the tank after 30 minutes: S(30) = -5e^(-30/50) + 6e^(30/50) ≈ 1.643 lb So, after 30 minutes, there will be approximately 1.643 pounds of salt in the tank.

answer:To solve the given differential equation dy/dx = x^2 + y^2, we can use the method of separation of variables. First, we rewrite the equation as: dy/(y^2) = dx/(x^2) Now, we integrate both sides: ∫(1/y^2) dy = ∫(1/x^2) dx Integrating both sides, we get: -1/y = -1/x + C Now, we solve for y: y = 1/(1/x + C) We can now use the initial condition y(0) = 1 to find the value of C: 1 = 1/(1/0 + C) Since 1/0 is undefined, we can rewrite the equation as: 1 = 1/C Thus, C = 1. Now, we have the equation: y = 1/(1/x + 1) To find the value of y when x = 1, we plug in x = 1: y = 1/(1/1 + 1) y = 1/(2) y = 1/2 So, the value of y when x = 1 is y = 1/2.