1)Given ODE ordinary differential equation of 1st order 1st degree:

y' + 2 y = Sin x

1) Generic solution: First we solve for y' + 2 y = 0 dy/dx = - 2 y dy/y = - 2 dx Ln y = -2 x + K y = e^{-2x + K} where K is some constant. or, y = A e⁻²ˣ , where A is some constant.

We can do this using the characteristic equation and using Euler method. y' + 2 y = 0 r + 2 = 0 => r = -2. Generic solution: c * e^{r x} = c e^{-2x}

2) To solve particular solution: y' + 2 y = Sin x Let y = a Sin x + b Cos x So y' = a Cos x - b Sin x

So a Cos x - b Sin x + 2 (a sin x + b cos x) = Sin xCompare the coefficients: 2a - b = 1 a + 2 b = 0Solving these equations we get: a = 2/5 and b = -1/5Particular solution: (2 Sinx - Cos x) / 5

Complete solution: y = A e⁻²ˣ + 2/5 Sin x - 1/5 * Cos x

4)i) Given one is a first order linear differential equation in the form of (dy/dx) + f(x)*y = g(x)Applying with a special method of using integrating factor this can be solved in a simple form.

ii) Here f(x) = sec(x); so ∫f(x) dx = ln|sec(x) + tan(x)|So the integrating factor is: [e^{∫f(x) dx}] = e^{ln|sec(x) + tan(x)|} = sec(x) + tan(x)

iii) Multiplying the given equation through out by the integrating factor,{sec(x) + tan(x)}*(dy/dx) + [sec(x){sec(x) + tan(x)}]*y = tan(x)*{sec(x) + tan(x)}

Integrating both sides, y*{sec(x) + tan(x)} = ∫{sec(x)*tan(x) + tan²x} -------- (1)

[Note: Derivative of y*{sec(x) + tan(x)} = (dy/dx)*{sec(x) + tan(x)} + {sec(x)*tan(x) + sec²x}*y]

iv) ∫{sec(x)*tan(x) + tan²x} dx = ∫{sec(x)*tan(x) + sec²x - 1} dx = sec(x) + tan(x) - x + C ----- (2)

Thus from (1) & (2):y*{sec(x) + tan(x)} = sec(x) + tan(x) - x + C