please solve the second order of differential equation probl

please solve the second order of differential equation problem (must show all work)

Solve the second order of differential equation problem d^2y/dx^2 minus y = e^2x

Solution

3 ) d²y / dx² - y = e^2x....................................................1

D-operator form is ,

   ( D² - 1 ) y = e^2x

The auxialary equation is ,

m² - 1 = 0

   m² - 1² = 0

( m + 1 ) ( m - 1 ) = 0 [ since , a² - b² = (a + b) (a-b) ]

m = -1 , 1

We know that

If the roots are real and imaginary then the solution is ,

y_c = c₁ e^mx + c₂ e^mx

Hence ,

y_c = c₁ e^1.x + c₂ e^-1.^x

  y_c = c₁ e^x + c₂ e^-x  

For a non homogeneous term e^2x we assume perticular solution is,

y_p = Ae^2x

Substitute y = Ae^2x in equation 1

d²y / dx² - y = e^2x

d²/dx²(Ae^2x ) -Ae^2x = e^2x............................2

d/dx( Ae^2x) = A.e^2x.2 = 2Ae^2x [since, d/dx(e^ax ) = ae^ax ]

   d²/dx²(Ae^2x ) = d/dx( 2Ae^2x )

= 2Ae^2x.2

= 4Ae^2x

From equation 2

  d²/dx²(Ae^2x ) -Ae^2x = e^2x

4Ae^2x - Ae^2x = e^2x

   3Ae^2x = e^2x

Equating the coefficients

3A = 1

A = 1/3

Hence,

   y_p = Ae^2x

   y_p = (1/3)e^2x

y_p = e^2x/3   

The general solution is ,

y(x) = y_c + y_p

= c₁ e^x + c₂ e^-x + e^2x/3   

Therefore,

The general solution is , y(x) = c₁ e^x + c₂ e^-x + e^2x/3