For the first equation: [ \frac{d^2y}{dx^2} - y = \frac{2}{1 + e^x} ] Step 1: Solve the complementary (homogeneous) equation [ \frac{d^2y}{dx^2} - y = 0 ] The characteristic equation is: [ r^2 - 1 = 0 \implies r = \pm 1 ] So the complementary solution is: [ y_c = C_1 e^x + C_2 e^{-x} ] Step 2: Use variation of parameters for the particular solution Let (y_p = u_1 e^x + u_2 e^{-x}), where [ u_1' e^x + u_2' e^{-x} = 0 \quad \text{and} \quad u_1' e^x - u_2' e^{-x} = \frac{2}{1 + e^x} ] From the first equation: [ u_1' e^x + u_2' e^{-x} = 0 \implies u_2' = -u_1' e^{2x} ] Substitute into the second equation: [ u_1' e^x - (-u_1' e^{2x}) e^{-x} = u_1' e^x + u_1' e^x = 2 u_1' e^x = \frac{2}{1 + e^x} \implies u_1' = \frac{1}{e^x (1 + e^x)} = \frac{1}{e^x + e^{2x}} = \frac{1}{e^x(1 + e^x)} ] [ u_1' = \frac{1}{e^x(1 + e^x)} = \frac{1}{e^x (1 + e^x)} ] [ u_1 = \int \frac{dx}{e^x (1 + e^x)} = \int \frac{e^{-x} dx}{1 + e^x} ] Let (t = e^x \implies dt = e^x dx \implies dx = \frac{dt}{t}) [ u_1 = \int \frac{e^{-x}}{1 + e^x} dx = \int \frac{1/t}{1 + t} \cdot \frac{dt}{t} = \int \frac{dt}{t^2(1+t)} ] Use partial fractions: [ \frac{1}{t^2 (1+t)} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{1+t} ] Multiply both sides by (t^2(1+t)): [ 1 = A t (1+t) + B (1+t) + C t^2 = A t^2 + A t + B + B t + C t^2 = (A + C) t^2 + (A + B) t + B ] Compare coefficients: [ t^2: A + C = 0 \implies C = -A ] [ t^1: A + B = 0 \implies B = -A ] [ t^0: B = 1 \implies B = 1 \implies -A = 1 \implies A = -1, C = 1 ] So [ \frac{1}{t^2 (1+t)} = -\frac{1}{t} + \frac{1}{t^2} + \frac{1}{1+t} ] Integrate term by term: [ u_1 = \int \left(-\frac{1}{t} + \frac{1}{t^2} + \frac{1}{1+t}\right) dt = -\ln|t| - \frac{1}{t} + \ln|1+t| = \ln\left|\frac{1+t}{t}\right| - \frac{1}{t} ] Back-substitute (t = e^x): [ u_1 = \ln\left(\frac{1+e^x}{e^x}\right) - e^{-x} = \ln(1+e^x) - x - e^{-x} ] [ u_2 = \int -u_1' e^{2x} dx = \int -\frac{1}{1 + e^x} e^x dx = -\int \frac{e^x}{1 + e^x} dx ] [ -\int \frac{e^x}{1+e^x} dx = -\ln|1 + e^x| ] Step 3: Particular solution [ y_p = u_1