Find the solution of the initial value problem by method of

Find the solution of the initial value problem, by method of undeterrmined coefficients,

y\'\'-2y\'-3y=3te^(2t) y(0)=1 y\'(0)=6

Solution

Given that

  y\'\'-2y\'-3y=3te^(2t) , y(0)=1 , y\'(0)=6

D-operator form is ,

d²y/dt² - 2dy/dt -3y = 3te^{(2t).....................................................1}

( D² - 2D -3 )y = 3te^(2t)

The auxialary equation is ,

m² - 2m - 3 = 0

( m - 3 ) ( m + 1 ) = 0

m = 3 , -1

The roots are real and imaginary

The complementary function is ,

y_c = c₁e^3t + c₂e^(-1)t

  y_c = c₁e^3t + c₂e^-t

For non homogeneous part g(t) =  3te^(2t),

let the perticular solution is , y_p = e^2t( At + B )

    d²/dt²( e^2t( At + B ) )  - 2d/dt ( e^2t( At + B ) ) - 3 ( e^2t( At + B ) ) = 3te^(2t)....................2

   d/dt( e^2t( At + B ) )  = A.d/dt( te^2t) + B d/dt( e^2t)

= A[ d/dt(t).e^2t + t.d/dt(e^2t) ]+ B. e^2t.2 [since, (f.g)\' = f\'.g\' , d/dt(e^ax)=ae^ax ]

= A[ 1.e^2t + t.e^2t.2 ] + 2Be^2t

   = Ae^2t + 2Ate^2t + 2Be^2t

d²/dt²( e^2t( At + B ) )  = d/dt ( Ae^2t + 2Ate^2t + 2Be^2t )  

   = A.d/dt(e^2t) +  2A.d/dt(te^2t) + 2B.d/dt(e^2t)

= A.e^2t.2 + 2A[ e^2t + 2t.e^2t ] + 2B.e^2t.2

= 2Ae^2t + 2Ae^2t +  2Ate^2t + 4Be^2t

= 4Ae^2t +  2Ate^2t + 4Be^2t

From 2

   d²/dt²( e^2t( At + B ) )  - 2d/dt ( e^2t( At + B ) ) - 3 ( e^2t( At + B ) ) = 3te^(2t)

     4Ae^2t +  2Ate^2t + 4Be^2t - 2 [ Ae^2t + 2Ate^2t + 2Be^2t ] - 3Ate^2t - 3Be^2t = 3te^(2t)

      4Ae^2t +  2Ate^2t + 4Be^2t - 2Ae^2t- 4Ate^2t - 4Be^2t - 3Ate^2t - 3Be^2t = 3te^(2t)

2Ae^2t - 3Ate^2t - 3Be^2t = 3te^(2t)

Equating the co-efficients

( 2A - 3B ) = 0

-3A = 3

A = -1

Substitute B in  ( 2A - 3B ) = 0

2(-1) - 3B = 0

-3B = 2

B = -2/3

Hence,

The perticular solution is , y_p = e^2t( At + B )

  y_p = e^2t( (-1)t + (-2/3) )

y_p = e^2t ( -t - (2/3) )

General solution is ,

y(t) = y_c + y_p

y(t) =c₁e^3t + c₂e^-t + e^2t ( -t - (2/3) )...............................3

y(0) = 1

From equation 3

y(t) =c₁e^3t + c₂e^-t + e^2t ( -t - (2/3) ).

  y(0) =c₁e³⁽⁰⁾ + c₂e^-0 + e²⁽⁰⁾ ( -0 - (2/3) ).

1 = c1 +c2 - (2/3)

c1 +c2 - (2/3) = 1 -> c₁ + c₂ = 1 +(2/3) = 5/3 ........................................4

y\'(0) = 6

y(t) =c₁e^3t + c₂e^-t - te^2t - (2/3)e^2t

y\'(t) = c₁e^3t.3 + c₂e^-t.(-1) - [ e^2t + 2t.e^2t ] - (2/3)e^2t.2

y\'(t) = 3c₁e^3t -  c₂e^-t - e^2t -  2t.e^2t - (4/3)e^2t

y\'(0) =3c₁e³⁽⁰⁾ -  c₂e^-(0) - e²⁽⁰⁾ -  2(0).e²⁽⁰⁾ - (4/3)e²⁽⁰⁾

6 = 3c₁ - c₂ - 1 - 0 - (4/3)

6 = 3c₁ - c₂ -1 - 4/3

3c₁ - c₂ = -7 - 4/3

3c₁ - c₂ = -25/3 ................................5

On solving equations 4 and 5

c₁ = -5/3 , c₂ = 10/3

Hence,

  y(t) =c₁e^3t + c₂e^-t + e^2t ( -t - (2/3) )

= -5/3e^3t + (10/3)e^-t +  e^2t ( -t - (2/3) )

Therefore,

General solution is ,  y(t) = (-5/3)e^3t + (10/3)e^-t +  e^2t ( -t - (2/3) )