Initial value problem D.E. (confused of a weird sol.)

[bobmerhebi]

Homework Statement

solve:

y" + 4y' + 4y = (3 + x)e^-2x ... (1); y(0) = 2 & y'(0) = 5

Homework Equations

using Undetermined Coefficients Method

The Attempt at a Solution

solving the associated homog. eq of (1) : y" + 4y' + 4y = 0 ... (2)

we get: y_c = c₁ e^-2x as m=-2

let f(x) = (3 + x)e^-2x

the UC set of :

a) 3e^-2x = {e^-2x} = S₁

b) xe^-2x = {xe^-2x , e^-2x} = S₂

as S₁ is included in S₂ then we omit S₁

but e^-2x is a solution of (2) as its in y_c then we multiply the 2ns set by x to get:

S'₂ = {x² e^-2x, xe^-2x } where both of the elements are NOT sol.'s of (2)

hence we get y_p = Axe^-2x + Bx² e^-2x

y'_p = Ae^-2x -2Axe^-2x + 2Bxe^-2x - 2Bx² e^-2x

& y"_p = -4Ae^-2x + 4Axe^-2x + 2Be^-2x - 8Bxe^-2x +4Bx² e^-2x

substitution in (1) one gets:

2Be^-2x = (3+x)e^-2x so B = 3+x/2 ? ??

where is A? why is B interms of x?

please help
thx

 

[gabbagabbahey]

solving the associated homog. eq of (1) : y" + 4y' + 4y = 0 ... (2)

we get: y_c = c₁ e^-2x as m=-2

Sure, m=-2 but with what multiplicity?...Remember, your homog DE is second order, so you expect two linearly independent solutions. You seem to only be using one of them.

This will also affect y_p.

 

[HallsofIvy]

Homework Statement

solve:

y" + 4y' + 4y = (3 + x)e^-2x ... (1); y(0) = 2 & y'(0) = 5

Homework Equations

using Undetermined Coefficients Method

The Attempt at a Solution

solving the associated homog. eq of (1) : y" + 4y' + 4y = 0 ... (2)

we get: y_c = c₁ e^-2x as m=-2

This is a second order equation. You need TWO independent solutions to form the general solution. Since m= -2 is a double root of the characteristic equation, xe^-x is the other, independent, solution.

let f(x) = (3 + x)e^-2x

the UC set of :

a) 3e^-2x = {e^-2x} = S₁

b) xe^-2x = {xe^-2x , e^-2x} = S₂

as S₁ is included in S₂ then we omit S₁

but e^-2x is a solution of (2) as its in y_c then we multiply the 2ns set by x to get:

S'₂ = {x² e^-2x, xe^-2x } where both of the elements are NOT sol.'s of (2)

hence we get y_p = Axe^-2x + Bx² e^-2x

y'_p = Ae^-2x -2Axe^-2x + 2Bxe^-2x - 2Bx² e^-2x

& y"_p = -4Ae^-2x + 4Axe^-2x + 2Be^-2x - 8Bxe^-2x +4Bx² e^-2x

substitution in (1) one gets:

2Be^-2x = (3+x)e^-2x so B = 3+x/2 ? ??

where is A? why is B interms of x?

please help
thx

A has disappeared and you can't solve for the constant B because both Ae^-2x and Bxe^-2x are already solutions to the homogeneous equation. In order not to have either e^-2x or xe^-2x you will need to multiply by x².

Try y= (Ax²+ Bx³) e^-2x and solve for A and B.

 

[bobmerhebi]

thx. i had in mind that i might have done a mistake in writing y_c but didn't bother to correct it. now i see how it caused the 2nd problem.
now after resolving it i got A =3/2 & B = 1/6

thus we now have:

y = c₁ e^-2x + c₂ xe^-2x + (3/2)x² e^-2x + (1/6)x³ e^-2x

solving for the conditions i got:

c₁ = 2 & c₂ = 9

thus y becomes:

y = 2e^-2x + 9xe^-2x + (3/2)x² e^-2x + (1/6)x³ e^-2x


i think this is the general solution. thanks 4 the help