y3y10y6e2t Solve the IVP y02 y00 y3y10y6e2t Solve the IVP y0

Solution

Given that

  y\"- 3y\' - 10y= 6e^(-2t) , y(0) = 2 , y\'(0) = 0

D-operator form of given equation is

( D² - 3D - 10 ) y = 6e^(-2t) .......................1

Auxialary equation is,

m² - 3m - 10 = 0

( m - 5 ) ( m + 2 ) = 0

m = 5 , -2

Hence,

The roots are real and distinct

complementary function y_c = c¹e^5t + c²e^-2t

For a non homogeneous part g(x) = 6.e^-2t , assume perticular solution y_p = Ate^-2t

   Substitute y_p in equation 1

  ( D² - 3D - 10 ) y = 6e^(-2t)

     ( D² - 3D - 10 ) Ate^-2t = 6e^(-2t)

D²( Ate^-2t ) - 3D( Ate^-2t) - 10Ate^-2t = 6e^(-2t) ............................2

D( Ate^-2t ) = A D( te^-2t )

= A [ d/d(t) .(e-2t) + t.d/dt(e-2t) ]

= A [ 1. (e-2t) + t . (e-2t).-2 ]

= A [ (e-2t) - 2t.(e-2t) ]

D²( Ate^-2t ) = D [ A [ (e^-2t) - 2t.(e^-2t) ]

= A D[ (e^-2t) - 2t.(e^-2t) ]

= A [ 4te^-2t - 4e^-2t ]

From 2

    D²( Ate^-2t ) - 3D( Ate^-2t) - 10Ate^-2t = 6e^(-2t)

  A [ 4te^-2t - 4e^-2t ] -3A [  (e^-2t) - 2t.(e^-2t) ] - 10Ate^-2t = 6e^(-2t)

4Ate^-2t - 4Ae^-2t - 3Ae^-2t + 6Ate^-2t -10Ae^-2t = 6e^(-2t)

   10Ae^-2t - 10Ae^-2t - 7Ae^-2t = 6e^(-2t)

  - 7Ae^-2t = 6e^(-2t)

   Equating the coefficients

-7A = 6

A = - 6 / 7

Hence,

Perticular solution y_p =  Ate^-2t

   y_p = (-6/7)te^-2t

General solution y(t) = y_c + y_p

y(t) = c₁e^5t + c₂e^-2t +(-6/7)te^-2t .......................................3

y(0) = 2 , i.e at t=0 y = 2

   y(0) = c₁e⁵⁽⁰⁾ + c₂e^-2(0) +(-6/7).0.e^-2(0)

   2 = c₁.1 + c₂.1 + 0

c₁ + c₂ = 2.....................................................4

y\'(0) = 0, i.e at t=0 y\' = 0

  y(t) = c₁e^5t + c2e^-2t +(-6/7)te^-2t

   y\'(t) = c₁.e^5t.5 + c₂.e^-2t.(-2) + (-6/7) [  (e^-2t) - 2t.(e^-2t) ]

y\'(t) = 5c₁e^5t -2c₂e^-2t + (-6/7) [  (e^-2t) - 2t.(e^-2t) ]

y\'(0) = 5c₁e⁵⁽⁰⁾ -2c₂e^-2(0) + (-6/7) [  (e^-2(0)) - 2(0) .(e^-2(0)) ]

0 = 5c1 - 2c2 - ( 6/7 )

5c1 - 2c2 = 6/7

35c1 - 14c2 = 6........................................................................5

solve equations 4 and 5

  c₁ + c₂ = 2

  35c₁ - 14c₂ = 6

On solving

c₁ = 34/49 , c₂ = 64/49

Substitute c1 ,c2 in equation 3

  y(t) = c₁e^5t + c₂e^-2t +(-6/7)te^-2t

   y(t) = (34/49)e_5t + (64/49)e^-2t - (6/7)te^-2t