Answer to Question #122591 in Differential Equations for jse

Question

1. Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficients. If possible, factor L. (Use D for the differential operator.)

y''' + 14y'' + 49y' = ex

_______ y= e^x

2. Find a linear differential operator that annihilates the given function. (Use D for the differential operator.)

1 + 3x − 4x3

Answer 1

Solution:

1. If Dy:=y' then D²y=y'',D³y=y''' and y''' + 14y'' + 49y' =D³y+14D²y+49Dy

We have equation D³y+14D²y+49Dy=e^x

and (D³+14D²+49D)y=e^x.

Then we can write the given equation in the form

L(y)=e^x with operator L=D³+14D²+49D

After factorization we obtain

D³+14D²+49D=D(D²+14D+49)=D(D+7)²

Answer:D(D+7)²=e^x

2.1+3x-4x³= (1+3x-4x³)e^0x

Polynom 1+3x-4x³ has degree 3.

Then $\lambda$ =0, k=4

Construct polynom which has root λ=0 of multiplicity 4:

λ⁴=0

Аnd appropriate linear differential operator with constant coefficients is L=D⁴

Answer:D⁴