The table that is needed has been provided in the images.Given the differential&nbsp;=y&nbsp;″−8y′+16y=−3e4x.a) Find the complementary solution&nbsp;=yc&nbsp;to the differential equation.b) Write down the FORM of the particular solution&nbsp;=yp&nbsp;for a solution using undetermined coefficients. DO NOT solve for&nbsp;=yp. Use Table. 4.4.1=yp=__________________________

Key to "The table that is needed has been provided..."

High School
Calculus and Analysis

Cristal Travis

Open question

2022-08-18

The table that is needed has been provided in the images.
Given the differential $= y^{″} - 8 y^{'} + 16 y = - 3 e^{4 x}$ .
a) Find the complementary solution $= y_{c}$ to the differential equation.
b) Write down the FORM of the particular solution $= y_{p}$ for a solution using undetermined coefficients. DO NOT solve for $= y_{p}$ . Use Table. 4.4.1
$= y_{p} =$ __________________________

Answer & Explanation

Bryanna Villarreal

Beginner2022-08-19Added 5 answers

Step 1
Given differential equation

= y^{″} - 8 y^{'} + 16 y = - 3 e^{4 x}

.
The auxiliary equation covespoding to homogen equation is

= m^{2} - 8 m + 16 = 0

Solve form

= {(m - 4)}^{2} = 0

=\Rightarrow m = 4, 4

The multiplicity of the root

m = 4

is 2 which gives

= y_{1} (x) = c_{1} e^{4 x}

and

= y_{2} (x) = c_{2} e^{4 x} \cdot x

Where

= c_{1}

and

c_{2}

are constant
The complementary solution is sum of the above solutions:

= y_{c} (x) = y_{1} (x) + y_{2} (x)

=\Rightarrow y_{c} (x) = c_{1} e^{4 x} + c_{2} e^{4 x} \cdot x

ureq8

Beginner2022-08-20Added 2 answers

Step 2
Now.
Defermine the particular solution to

y^{″} - 8^{'} + 16 y = - 3 e^{4 x}

. by the method of undertemined coefficietns:
The particular solution to

= y^{″} - 8^{'} + 16 y = - 3 e^{4 x}

is of the form

= y_{p} (x) = x^{2} (a_{1} e^{4 x})

where

= a_{1}

where

a_{1} e^{4 x}

was multiplied by

x^{2}

to account for

e^{4 x} x

in the complementary solution.

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