Question

dy/dx+2y=sinx. Find the general solution

Answer 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 x
Compare the coefficients:
       2a - b = 1
       a + 2 b = 0
Solving these equations we get:   a = 2/5   and  b = -1/5
Particular solution:   (2 Sinx - Cos x) / 5

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

Answer 2

Answer:

y'+2y=sinx

Step-by-step explanation:

dy/dx=-2y