Question

Solve `(d^2y)/(dx^2) -4 dy/dx + 3y = e^xcos(2x)` ?

Answer 1

See below.

Explanation:

Considering `D equiv d/(dx)` and applying the operator

`F(s) = L(f(x))=int_0^oo e^(-sx)f(x)dx` and considering null initial conditions, we have

`(s^2-4s+3)Y(s) = int_0^oo e^(-sx)e^x cos(2x)dx =(s-1)/(4 + (s-1)^2)` or

`Y(s) = (s-1)/((s^2-4s+3)(4 + (s-1)^2) ) = a/(s-1)+b/(s-3)+(cs+d)/(s^2-2s+5)`

and inverting

`y(x)=e^x (a + b e^(2 x) + c cos(2x) + (c + d) cosx sinx)`

NOTE:
`a = lim_(s->1)(s-1)((s-1)/((s^2-4s+3)(4 + (s-1)^2) ))=0`
`b = lim_(s->3)(s-3)((s-1)/((s^2-4s+3)(4 + (s-1)^2) ))=1/8`

etc.

Answer 2

`y(x) = Ae^(x) + Be^(3x) -1/8e^xcos2x-1/8e^xsin2x`

Explanation:

We have:

`(D^2-4D+3)y=xe^x`

Where `D` is the linear differential operator `d/dx`. Thus we can write the equation as:

`(d^2y)/(dx^2) -4 dy/dx + 3y = e^xcos(2x)` ..... [A]

This is a second order linear non-Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution, `y_c` of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, `y_p` of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

`(d^2y)/(dx^2) -4dy/dx + 3y = 0`

And it's associated Auxiliary equation is:

`m^2 -4m+3 = 0 => (m-3)(m-1) = 0`

Which has two real and distinct solutions `m=1,3`

Thus the solution of the homogeneous equation is:

`y_c = Ae^(1x) + Be^(3x)`
`\ \ \ = Ae^(x) + Be^(3x)`

Particular Solution

In order to find a particular solution of the non-homogeneous equation we would look for a solution of the form:

`y = (acos2x+bsin2x)e^x`

Where the constants `a` and `b` are to be determined by direct substitution and comparison:

Differentiating wrt `x` (using the product rule) we get:

`dy/dx = (acos2x+bsin2x)(e^x) + (-2asin2x+2bcos2x)(e^x)`
`" " = ((a+2b)cos2x+(b-2a)sin2x)(e^x)`

Differentiating again wrt `x` (using the product rule) we get:

`(d^2y)/(dx^2) = ((a+2b)cos2x+(b-2a)sin2x)(e^x)+(-2a-4b)sin2x+(2b-4a)cos2x)(e^x)`
`" " = ((a+2b +2b-4a )cos2x+ (b-2a-2a-4b)sin2x)(e^x)`
`" " = ((-3a+4b)cos2x+ (-4a-3b)sin2x)(e^x)`

Substituting into the DE [A] we get:

`((-3a+4b)cos2x+ (-4a-3b)sin2x)(e^x) -4 ((a+2b)cos2x+(b-2a)sin2x)(e^x) + 3(acos2x+bsin2x)e^x = e^xcos(2x)`

`:. (-3a+4b -4 (a+2b) + 3a)cos2x+ (-4a-3b-4(b-2a)+3b)sin2x = cos(2x)`
`:. (-4a-4b)cos2x + (4a-4b)sin2x = cos(2x)`

Equating coefficients of `cos2x` and `sin2x` we get

`cos2x: -4a-4b=1`
`sin2x: 4a-4b = 0`

Solving simultaneously, we have:

`a = -1/8`
`b = -1/8`

And so we form the Particular solution:

`y_p = (-1/8cos2x-1/8sin2x)e^x`
`\ \ \ = -1/8e^xcos2x-1/8e^xsin2x`

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Ae^(x) + Be^(3x) -1/8e^xcos2x-1/8e^xsin2x`