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Sagot :
Particular solution for the equation a. Y" – 6y + 5y = 5x2 + 3x – 16 – 9223 is
5y = -15e²⁷ - 23/5
Particular solution for b. y" – 6y' + 5y = -10x2 - 6x + 32 +622 is 5y = 8x+ 218
Given:
y= 3e²⁷ (Solution 1)
Y = x² + 3x (Solution 2)
y" - 6y' + 5y = -9222 (Equation 1)
y" - 6y' + 5y = 5x² + 3x – 16 (Equation 2)
Where [tex]y^{"}[/tex] represents the double derivative(d²y) of y and [tex]y^{'}[/tex] represents the first derivative(dy) of the equation y.
Find Solutions for:
a. Y" – 6y + 5y = 5x² + 3x – 16 – 9223
b. y" – 6y' + 5y = -10x² - 6x + 32 +622
y= 3e²⁷ (Solution 1)(Given in the question)
Taking derivative on both sides
by = 3 de²⁷
∵ e²⁷ is constant in comparison to dy the derivative will be 0. Thus, d²y of y is also equal to zero
Then, putting values in the equation 1
- 6y' + 5y + y" = -9222 (Equation 1)
(0) - 6(0) +5(3e²⁷) = -9222
Thus, 15e²⁷ = -9222
5e²⁷ = -3074
e²⁷ = -614.8
Y = x² + 3x (Solution 2)(Given in the question)
Taking derivative on both sides
dY/dx= 2x + 3 -(iii)
taking derivative of y'
d²Y/dx² = 2
∵ 3 is a constant in dy/dx.
Then, putting values in the equation 2
y" -+ 5y - 6y' = 5x² + 3x - 16
(2) - 6(2x + 3) +5y = 5x² + 3x - 16 + 12x - 12x
2 - 12 + 18x + 5y = 5x² +15x - 16 -12x
-10 + 18x - 15x + 5y = 5y -16 -12x
3x + 12x =10 - 16
15x = -6
x= -2/5
Putting values of x, y, dy and d²y in equations (i) & (ii)
a. Y" – 6y + 5y = 5x² + 3x - 16 – 9223
(2) -6(2x+3) + 5y = 5x² + 3x - 16 - 9223 + 12x - 12x
2 - 12x -18 + 5y = 5y - 9223 - 16 - 12x
2 + - 18 + 5y -6/5 -(9222 + 1)
2 + 24-6/5 + 5y + 15e²⁷ - 1 = 0
5y + 15e²⁷ + 18/5 + 1= 0
5y + 15e²⁷ + 23/5= 0
5y = -15e²⁷ - 23/5
b. y" – 6y' + 5y = -10x² - 6x + 32 +622
(0) -6(0) + 5y = -10x² - 6x +654
5y = -10x² - 6x -24x + 24x + 654
5y = -10(x² + 3x) +24x + 654
5y = -10y + 24x + 654
5y +10y = 24x + 654
15y=24x+654
5y = 8x+ 218
Learn more about derivatives at:
brainly.com/question/23847661
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