Answer:
(a) The particular solution, y_p is 7
(b) y_p is -4x
(c) y_p is -4x + 7
(d) y_p is 8x + (7/2)
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 = 14.
Because the nonhomogeneus part of the differential equation, 14 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 = 14
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 = 14
A = 6/2 = 7
Therefore, the particular solution, y_p = A is 7
(b) y'' + 2y = −8x
Let y_p = Ax + B
y'_p = A
y''_p = 0
0 + 2(Ax + B) = -8x
2Ax + 2B = -8x
By inspection,
2B = 0 => B = 0
2A = -8 => A = -8/2 = -4
The particular solution y_p = Ax + B
is -4x
(c) y'' + 2y = −8x + 14
Let y_p = Ax + B
y'_p = A
y''_p = 0
0 + 2(Ax + B) = -8x + 14
2Ax + 2B = -8x + 14
By inspection,
2B = 14 => B = 14/2 = 7
2A = -8 => A = -8/2 = -4
The particular solution y_p = Ax + B
is -4x + 7
(d) Find a particular solution of y'' + 2y = 16x + 7
Let y_p = Ax + B
y'_p = A
y''_p = 0
0 + 2(Ax + B) = 16x + 7
2Ax + 2B = 16x + 7
By inspection,
2B = 7 => B = 7/2
2A = 16 => A = 16/2 = 8
The particular solution y_p = Ax + B
is 8x + (7/2)
