Respuesta :
Answer:
(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
