The critical numbers of the function f(x) = x⁶ - x⁴ + 2x³ - 3x.
x₅ = 1.35240 is correct to six decimal places.
Use Newton's method to find the critical numbers of the function
Newton's method
[tex]x_{x+1} = x_n - \frac{x_n^6-(x_n)^4+2(x_n)^3-3x}{6(x_n)^5-4(x_n)^3+6(x_n)-3}[/tex]
f(x) = x⁶ - x⁴ + 2x³ - 3x
f'(x) = 6x⁵ - 4x³ + 6x² - 3
Now plug n = 1 in equation
[tex]x_{1+1} = x_n -\frac{x^6-x^4+2x^3=3x}{6x^5-4x^3+6x^2-3} = \frac{6}{5}[/tex]
Now, when x₂ = 6/5, x₃ = 1.1437
When, x₃ = 1.1437, x₄ = 1.135 and when x₄ = 1.1437 then x₅ = 1.35240.
x₅ = 1.35240 is correct to six decimal places.
Therefore, x₅ = 1.35240 is correct to six decimal places.
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Find the general solution to the following differential equations:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x^2
The general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
Given differential equations are:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x²
To find the general solution to the given differential equations, we will solve these equations one by one.
(i) 16y'' - 8y' + y = 0
The characteristic equation is:
16m² - 8m + 1 = 0
Solving this quadratic equation, we get m = 1/4, 1/4
Hence, the general solution of the given differential equation is:
y = c₁e^(x/4) + c₂xe^(x/4)..................................................(1)
(ii) y" + y' - 2y = 0
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2
Hence, the general solution of the given differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(2)
(iii) y" + y' - 2y = x²
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2.
The complementary function (CF) of this differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(3)
Now, we will find the particular integral (PI). Let's assume that the PI of the differential equation is of the form:
y = Ax² + Bx + C
Substituting the value of y in the given differential equation, we get:
2A - 4A + 2Ax² + 4Ax - 2Ax² = x²
Equating the coefficients of x², x, and the constant terms on both sides, we get:
2A - 2A = 1,
4A - 4A = 0, and
2A = 0
Solving these equations, we get
A = 1/2,
B = 0, and
C = 0
Hence, the particular integral of the given differential equation is:
y = (1/2)x²..................................................(4)
The general solution of the given differential equation is the sum of CF and PI.
Hence, the general solution is:
y = c₁e^x + c₂e^(-2x) + (1/2)x²..................................................(5)
Conclusion: Therefore, the general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
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The particular solution is: y = -1/2 x². The general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
The general solution of the given differential equations are:
Given differential equation: 16y'' - 8y' + y = 0
The auxiliary equation is: 16m² - 8m + 1 = 0
On solving the above quadratic equation, we get:
m = 1/4, 1/4
∴ General solution of the given differential equation is:
y = c1 e^(x/4) + c2 x e^(x/4)
Given differential equation: y" + y' - 2y = 0
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:
m = -2, 1
∴ General solution of the given differential equation is:
y = c1 e^(-2x) + c2 e^(x)
Given differential equation: y" + y' - 2y = x²
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:m = -2, 1
∴ The complementary solution is:y = c1 e^(-2x) + c2 e^(x)
Now we have to find the particular solution, let us assume the particular solution of the given differential equation:
y = ax² + bx + c
We will use the method of undetermined coefficients.
Substituting y in the differential equation:y" + y' - 2y = x²a(2) + 2a + b - 2ax² - 2bx - 2c = x²
Comparing the coefficients of x² on both sides, we get:-2a = 1
∴ a = -1/2
Comparing the coefficients of x on both sides, we get:-2b = 0 ∴ b = 0
Comparing the constant terms on both sides, we get:2c = 0 ∴ c = 0
Thus, the particular solution is: y = -1/2 x²
Now, the general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
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