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3 years ago

What are the non permissible values of (2n-5)/(3n+1)

Mathematics

1 answer:

myrzilka [38]3 years ago

6 0

Answer:

n= -1/3

Step-by-step explanation:

To find non permissible values (npv), focus on the denominator of an expression

In this case, the denominator is: (3n+1)

The npv is basically the value that would make the expression equal to 0

So, make the expression in the denominator equal 0

3n+1 = 0

Solve for n

3n= -1

n = (-1/3)

Thus, the npv for this equation is n= -1/3

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Consider the equation

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Answer:

A. $a=10\\ \\b\neq -8$

B. $a=10\\ \\b= -8$

Step-by-step explanation:

Consider the equation

$2(5x-4) = ax + b$

A. This equation has no solutions when the coefficients at x are the same and the free coefficients are not the same.

First, use distributive property:

$2(5x-4)=2\cdot 5x-2\cdot 4=10x-8$

So, the equation is

$10x-8=ax+b$

This equation has no solutions when

$a=10\\ \\b\neq -8$

B. The equation has infinitely many solutions when the coefficients at x are the same and the free coefficients are the same too.

So, the equation

$10x-8=ax+b$

has infinitely many solutions when

$a=10\\ \\b= -8$

In other cases, the equation has a unique solution

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Step-by-step explanation:

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3 years ago

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

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