Word Count : 1055

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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 __________ ____ ___ ____ ______ __________ _____ ___.
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