Mathematics, 27.02.2020 19:54 brien301

Consider the differential equationy '' − 2y ' + 26y = 0; ex cos 5x, ex sin 5x, (−[infinity], [infinity]).Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. The functions satisfy the differential equation and are linearly independent sinceW(ex cos 5x, ex sin 5x) = ≠ 0for−[infinity] < x < [infinity].

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Answer from: mez59

Both given solutions form a fundamental set of solutions, since they are solutions of the given differential equation and their Wronskian is $W=5e^{2x}$ which is different than 0, thus the solutions are linearly independent.

Step-by-step explanation:

In order to verify that the given functions form a fundamental set of solutions of the differential equation, we need to replace them on the differential equation. After we have verified both solutions, we can find the Wronskian to show that they are linearly independent.

Verifying the given functions as solutions of the given ODE.

For $y = e^x \cos (5x)$ we can find the first derivative using product rule.

$y '= -e^x \sin (5x)(5)+\cos(5x)e^x \\ y'=-5 e^x \sin (5x) +e^x \cos(5x)$

And the second derivative will be

$y'' =-25 e^x \cos(5x) -5 e^x \sin (5x) -5e^x \sin(5x)+e^x \cos(5x)\\ y''=-10 e^x \sin (5x) -24e^x \cos(5x)$

Replacing on the given ODE we get

y'' -2y'+26y=0

$-10 e^x \sin (5x) -24e^x \cos(5x)-2(-5 e^x \sin (5x) +e^x \cos(5x))+26(e^x \cos(5x))=0$

$-10 e^x \sin (5x) -24e^x \cos(5x)+10 e^x \sin (5x) -2e^x \cos(5x)+26e^x \cos(5x)=0$

Combining like terms.

0=0

We can repeat the process for $y = e^x \sin(5x)$ , so using product rule we have

$y '= e^x \cos (5x)(5)+\sin(5x)e^x \\ y'=5 e^x \cos (5x) +e^x \sin(5x)$

And the second derivative is

$y''=-25 e^x \sin (5x) +5 e^x \cos (5x) +5e^x \cos(5x)+e^x \sin(5x)\\y''=-24 e^x \sin (5x)+10e^x \cos(5x)$

Replacing on the given ODE we get

$y''-2y'+26y=0\\ -24 e^x \sin (5x)+10e^x \cos(5x)-2(5e^x \cos(5x)+e^x \sin(5x)) +26e^x \sin(5x)=0$

$-24 e^x \sin (5x)+10e^x \cos(5x)-10e^x \cos(5x)-2e^x \sin(5x) +26e^x \sin(5x)=0$

Combining like terms we get

0=0

Thus we have verified that the functions are solutions of the differential equation on the indicated interval.

Verifying that the solutions are linearly independent.

We can use the first derivatives to find the Wronskian

$W(e^x \cos(5x), e^x \sin (5x)) =\left|\begin{array}{cc}e^x \cos(5x)&e^x \sin(5x)\\-5e^x \sin(5x)+e^x \cos(5x)&5e^x \cos(5x)+e^x \sin(5x)\end{array}\right|$ So we get

$W=e^x \cos(5x)(5e^x \cos(5x)+e^x \sin(5x))-e^x \sin(5x)(-5e^x \sin(5x)+e^x \cos(5x)) \\W=5e^{2x} \cos^2(5x)+e^{2x} \cos(5x)\sin(5x)+5e^{2x}\sin^2(5x)-e^{2x} \cos(5x)\sin(5x)$

Simplifying we get

$W=5e^{2x} \cos^2(5x)+5e^{2x}\sin^2(5x) \\W=5e^{2x}(\cos^2(5x)+\sin^2(5x))$

Using Pythagorean identity we get

$W=5e^{2x} \ne 0$

So it is not 0, thus we have verified that the solutions are linearly independent, thus they form a fundamental set of solutions.

Answer from: Quest

send it over

step-by-step explanation:

Answer from: Quest

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.

Which construction does the image below demonstrate? 1. an equilateral triangle circumscribed abou

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