A. By inspection, find a particular solution of y'' + 2y = 6. yp(x) =
(b) By inspection, find a particular solution of y'' + 2y = −12x. yp(x) =
(c) Find a particular solution of y'' + 2y = −12x + 6. yp(x) =
(d) Find a particular solution of y'' + 2y = 24x + 3. yp(x) =

(a) The particular solution, y_p is 3

(b) y_p is -6x

(c) y_p is -6x + 3

(d) y_p is 12x + 3

Step-by-step explanation:

To find a particular solution to a differential equation by inspection - is to assume a trial function that looks like the nonhomogeneous part of the differential equation.

(a) Given y'' + 2y = 6.

Because the nonhomogeneus part of the differential equation, 6 is a constant, our trial function will be a constant too.

Let A be our trial function:

We need our trial differential equation y''_p + 2y_p = 6

Now, we differentiate y_p = A twice, to obtain y'_p and y''_p that will be substituted into the differential equation.

y'_p = 0

y''_p = 0

Substitution into the trial differential equation, we have.

0 + 2A = 6

A = 6/2 = 3

Therefore, the particular solution, y_p = A is 3

(b) y'' + 2y = −12x

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -12x

2Ax + 2B = -12x

By inspection,

2B = 0 => B = 0

2A = -12 => A = -12/2 = -6

The particular solution y_p = Ax + B

is -6x

(c) y'' + 2y = −12x + 6

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -12x + 6

2Ax + 2B = -12x + 6

By inspection,

2B = 6 => B = 6/2 = 3

2A = -12 => A = -12/2 = -6

The particular solution y_p = Ax + B

is -6x + 3

(d) Find a particular solution of y'' + 2y = 24x + 3

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = 24x + 6

2Ax + 2B = 24x + 6

By inspection,

2B = 6 => B = 6/2 = 3

2A = 24 => A = 24/2 = 12

The particular solution y_p = Ax + B

is 12x + 3

b

step-by-step explanation:

given

- 2 (3x + 12y - 5 - 17x - 16y + 4 )

simplify the parenthesis by collecting like terms

= - 2 (- 14x - 4y - 1) ← distribute the parenthesis by - 2

= 28x + 8y + 2 → b