How to solve y ′ + 6 y ( t ) + 9 ∫ 0 t y ( τ ) d τ = 1, y ( 0 ) = 0

Question

How to solve       y    ′    +  6  y  (  t  )  +  9      ∫    0    t    y  (  τ  )  d  τ  =  1,     y  (  0  )  =  0

Beckham Krueger · Accepted Answer

Let  z  (  t  )  =      ∫    0    t    y  (  τ  )  d  τthen the ODE becomes      z    ″    +  6      z    ′    +  9  z  =  1  ,        z    ′    (  0  )  =  0Notice that       z    p    =      1    9   is a particular solution and for the homogenous solution the characteristic polynomial is      r    2    +  6  r  +  9  =  (  r  +  3      )    2  and r=−3 is the unique solution so      z    h    (  t  )  =  (  a  t  +  b  )      e          −      3      t      is the solution for the homogenous equation, and finally the general solution is  z  (  t  )  =      z    h    (  t  )  +      z    p  The coefficients a and b are determined by z(0)=0 and z′(0)=0 and differentiate z to find y.

How to solve y′+6y(t)+9 int_0^t y(tau)d tau=1, y(0)=0

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