Calculate the particular solution of dy/dx - y/(x ln x) = 1/x at (2; 0). Determine the general solution of d2y/dx2 + dy/dx - 6y = 2e^-3x.

Here are the solutions to the given differential equations. 4.1 Calculate the particular solution of (dy)/(dx) - (y)/(x x) = (1)/(x) at (2, 0). Step 1: Rewrite the differential equation in the standard form for a first-order linear differential equation, (dy)/(dx) + P(x)y = Q(x). (dy)/(dx) + (-(1)/(x x))y = (1)/(x) Here, P(x) = -(1)/(x x) and Q(x) = (1)/(x). Step 2: Find the integrating factor, I(x) = e^ P(x) dx. First, calculate P(x) dx: -(1)/(x x) dx Let u = x, then du = (1)/(x) dx. -(1)/(u) du = -|u| = -| x| Now, find the integrating factor: I(x) = e^-| x| = e^(( x)^-1) = (1)/( x) (Assuming x>1 for x to be positive and defined, which is true for x=2). Step 3: Multiply the differential equation by the integrating factor and integrate. The left side becomes (d)/(dx)(y · I(x)). (d)/(dx)(y · (1)/( x)) = (1)/(x) · (1)/( x) (d)/(dx)((y)/( x)) = (1)/(x x) Integrate both sides with respect to x: (d)/(dx)((y)/( x)) dx = (1)/(x x) dx (y)/( x) = | x| + C The general solution is y = x (| x| + C). Step 4: Use the initial condition (2, 0) to find the particular solution. Substitute x=2 and y=0 into the general solution: 0 = 2 (| 2| + C) Since 2 ≠ 0, we must have: | 2| + C = 0 C = -| 2| Substitute the value of C back into the general solution: y = x (| x| - | 2|) Using the logarithm property a - b = ((a)/(b)): y = x ( | ( x)/( 2) | ) The particular solution is: y = x (( x)/( 2)) 4.2 Determine the general solution of (d^2 y)/(dx^2) + (dy)/(dx) - 6y = 2e^-3x. Step 1: Find the complementary solution y_c by solving the homogeneous equation (d^2 y)/(dx^2) + (dy)/(dx) - 6y = 0. The characteristic equation is: r^2 + r - 6 = 0 Factor the quadratic equation: (r+3)(r-2) = 0 The roots are r_1 = -3 and r_2 = 2. The complementary solution is: y_c = C_1 e^-3x + C_2 e^2x Step 2: Find the particular solution y_p using the method of undetermined coefficients. The right-hand side is g(x) = 2e^-3x. Since -3 is a root of the characteristic equation (and it's a simple root), we must multiply the standard guess by x. Let y_p = Ax e^-3x. Calculate the first and second derivatives of y_p: y_p' = A e^-3x + Ax(-3e^-3x) = A e^-3x - 3Ax e^-3x = A(1-3x)e^-3x y_p'' = A(-3e^-3x) + (-3A e^-3x - 3Ax(-3e^-3x)) = -3A e^-3x - 3A e^-3x + 9Ax e^-3x = A(-6+9x)e^-3x Substitute y_p, y_p', and y_p'' into the non-homogeneous differential equation: A(-6+9x)e^-3x + A(1-3x)e^-3x - 6(Ax e^-3x) = 2e^-3x Divide by e^-3x: A(-6+9x) + A(1-3x) - 6Ax = 2 -6A + 9Ax + A - 3Ax - 6Ax = 2 Combine like terms: (-6A + A) + (9Ax - 3Ax - 6Ax) = 2 -5A + 0x = 2 -5A = 2 A = -(2)/(5) So, the particular solution is: y_p = -(2)/(5) x e^-3x Step 3: Write the general solution y = y_c + y_p. y = C_1 e^-3x + C_2 e^2x - (2)/(5) x e^-3x The general solution is: y = C_1 e^-3x + C_2 e^2x - (2)/(5) x e^-3x Send me the next one 📸