Suppose that a fourth order differential equation has a solution y = -7e5x x sin

Question

Suppose that a fourth order differential equation has a solution y = -7e5x x sin(x). Find such a differential equation, assuming it is homogeneous and has constant coefficients. y=1/(y -20y +152y''-520y'+676) help (equations) Note: for part (a), use primes to denote derivatives. Find the general solution to this differential equation. In your answer, use A, B, C and D to denote arbitrary constants and x the independent variable. y=Aexp(5x)sinx+Bexp(5x)cosx+Cxexp(5x)sinx+Dxexp(5x)cosx help (equations) Note: for this part, you MUST enter y = before your answer. For example, if I thought the answer was y = Aex + Bx2 I would literally enter y = Aex + Bx2.

Explanation / Answer

y' (x) = -35x.e^(5x) .sin(x) - 7e^(5x) sin(x) - 7x e^(5x) cos(x)

y''(x)= -168x. e^(5x).sin(x) - 70e^(5x) . sin(x) - 70x.e^(5x).cos(x) - 14e^(5x).cos(x)

y'''(x) = -770x.e^(5x).sin (x) - 504e^(5x). sin(x) - 518x.e^(5x). cos(x) - 210e^(5x).cos(x)

y''''(x) = -3332x.e^(5x).sin(x) - 3080e^(5x).sin(x) - 3360x.e^(5x).cos(x) - 2072e^(5x).cos(x)

the general form of the solution will be:

Ay'''' + by''' + cy'' + dy' + y = 0

We have to solve the following equations:

3332a + 770b + 168 c + 35 d = -7

3080a + 504b + 70c + 7d = 0

3360a + 518b + 70c + 7d = 0

2072a + 210 b + 14c = 0

Eliminating one of the variables we get:

280a + 14b = 0

12068a + 1750b + 182 c = 35

2072a + 210b + 14 c = 0

a = 5/676

b = -25/169

c = 190/169

d= -50/13

Thus, the equation is :

5y'''' - 100y''' + 760y'' - 2600 y' + 676y =0

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