Question

(d^2-4d-5)y=e^2x+3 cos(4x+3)​

Answer 1

(D+1)(D+3)y=e

−x

sinx+xe

3x

C.F. = C

1

e

−x

+C

2

e

−3x

For P.I. , we get

D

2

+4D+3

e

−x

sinx+xe

3x

e

−x

D(D+2)

sinx

+e

3x

(D+4)(D+6)

x

e

−x

D(D

2

−4)

(D−2)sinx

+e

3x

24

24

(D

2

+10D)

+1

x

=e

−x

−3

(sinx+2cosx)

+e

3x

288

5

Answer 2

The particular solution is:

$y_p = (5/4)e^(2x) + (1/16)sin(4x)$

To solve the differential equation:

$(d^2 - 4d - 5)y = e^(2x) + 3cos(4x + 3)$

We first find the characteristic equation by assuming that the solution is of the form $y = e^(rx):$

$r^2 - 4r - 5 = 0$

Solving this quadratic equation, we get:

r = (4 ± sqrt(16 + 4*5)) / 2

 = (4 ± sqrt(36)) / 2

 = 2 ± 3

So the roots are r = -1 and r = 5. Therefore, the general solution to the homogeneous equation $(d^2 - 4d - 5)y = 0$ is:

$y_h = c_1e^(-x) + c_2e^(5x)$

where $c_1 and c_2$ are constants.

Next, we find a particular solution to the non-homogeneous equation. Since the right-hand side contains both e^(2x) and cos(4x + 3), we try a particular solution of the form:

$y_p = Ae^(2x) + Bcos(4x) + Csin(4x)$

where A, B, and C are constants to be determined. Taking the first and second derivatives o f$y_p$, we get:

$y_p' = 2Ae^(2x) - 4Bsin(4x) + 4Ccos(4x)$

$y_p'' = 4Ae^(2x) - 16Bcos(4x) - 16Csin(4x)$

Substituting these into the original differential equation and simplifying, we get:

$(4A - 16B - 16C)e^(2x) + (48C)cos(4x) + (-24B)sin(4x) = e^(2x) + 3cos(4x + 3)$

To satisfy this equation, we must have:

4A - 16B - 16C = 1

48C = 3

-24B = 0

Solving for A, B, and C, we get:

A = 5/4, B = 0, C = 1/16

Therefore, the particular solution is:

$y_p = (5/4)e^(2x) + (1/16)sin(4x)$

The general solution to the non-homogeneous equation is then:

$y = y_h + y_p = c_1e^(-x) + c_2e^(5x) + (5/4)e^(2x) + (1/16)sin(4x)$

where$c_1 and c_2$ are constants determined by any initial or boundary conditions.

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