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tester [92]
3 years ago
9

Solve this problem by using variation of parameters method. y''-y=coshx.

Mathematics
1 answer:
Gekata [30.6K]3 years ago
3 0
y''-y=0\implies r^2-1=0\implies r=\pm1
\implies y_c=C_1e^x+C_2e^{-x}={C^*}_1\underbrace{\cosh x}_{y_1}+{C^*}_2\underbrace{\sinh x}_{y_2}

For the nonhomogeneous ODE

y''-y=\underbrace{\cosh x}_{f(x)}

we're looking for a particular solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\int\frac{y_2(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the two fundamental solutions.

We have

W(y_1,y_2)=\begin{vmatrix}\cosh x&\sinh x\\\sinh x&\cosh x\end{vmatrix}=\cosh^2x-\sinh^2x=1

so we're left with

u_1=-\displaystyle\int\sinh x\cosh x\,\mathrm dx=-\dfrac12\cosh^2x
u_2=\displaystyle\int\cosh^2x\,\mathrm dx=\dfrac12x+\dfrac14\sinh2x

so that the particular solution is

y_p=-\dfrac12\cosh^3x+\dfrac12x\sinh x+\dfrac14\sinh x\sinh2x
y_p=-\dfrac12\cosh x+\dfrac12x\sinh x

As y_1 already accounts for the \cosh x term in y_p, we're left with the general solution

y=C_1\cosh x+C_2\sinh x+\dfrac12x\sinh x
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Marina86 [1]

The zero of the function is at 33.69 degree , the graph is plotted and attached with the answer.

<h3>What is a Function ?</h3>

A function is a law that relates a dependent variable and an independent variable with each other

It is given that

y = 2tan (x - π/2) +3

To find the zeroes of a function that function has to be equated to zero.

2tan (x - π/2) +3 = 0

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x - 90 = -56.31

x = 33.69 degree

The zero of the function is at 33.69 degree

For finding the maxima /minima

the derivative is

dy/dx = 2 sec² (x - π/2)

the point at which the slope is zero is substituted in the second derivative to find maxima/minima

d²y/dx² = 4 sec² (x - π/2) tan (x - π/2)

if the value is negative then it is a maxima and if it is positive it is a minima.

The vertical asymptote is found by finding the values that make the function undefined

x = 0+ πn

No horizontal or oblique asymptote

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Answer:

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A game is played using ne die. If the die is rolled and shows 3, the player wins $45. If the die shows any number other than 3,
I am Lyosha [343]

Answer:

-$13.5

Step-by-step explanation:

Let x be a random variable of a count of player gain.

- We are told that if the die shows 3, the player wins $45.

- there is a charge of $9 to play the game

If he wins, he gains; 45 - 9 = $36

If he looses, he has a net gain which is a loss = -$9

Thus, the x-values are; (36, -9)

Probability of getting a 3 which is a win is P(X) = 1/6 since there are 6 numbers on the dice and probability of getting any other number is P(X) = 5/6

Thus;

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E(X) = (1/6)(36 - (5 × 9))

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E(X) = -3/2

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