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Answer» 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. hit like if you find it useful |
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