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kolezko [41]
1 year ago
15

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

a second solution y2(x) of the homogeneous equation and a particular solution yp(x) of the given nonhomogeneous equation. y'' − 7y' + 6y = x; y1 = ex

Mathematics
1 answer:
fenix001 [56]1 year ago
6 0

Answer:

y2 = (6x + 7)/36 + (Dx + E)e^x

Step-by-step explanation:

The method of reduction of order is applicable for second-order differential equations.

For a known solution y1 of a 2nd order differential equation, this method assumes a second solution in the form Uy1 which satisfies the said differential equation. It then assumes a reduced order for U'' (w' = U'').

The differential equation becomes easy to solve, and all that is left are integration and substitutions.

Check attachments for the solution to this problem.

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This figure has two intersecting lines and a ray. What is the value of x?
Evgen [1.6K]
The answer is 37 degrees
7 0
1 year ago
The length of a rectangle is measured as 370mm correct to 2 significant figures.
olga nikolaevna [1]

Answer:

UB: 375

Step-by-step explanation:

370mm to 2 sig figs..

UB: 375

LB:365

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5 0
2 years ago
Water leaves a spigot at a rate of 462 cubic inches per minute. How many cubic feet of water is this per hour? (Round your answe
irina [24]

Answer:

16 cubic feet per hour

Step-by-step explanation:

<em>This is simple conversion from inches to feet.</em>

<em>Step 1: Write the scale of conversion</em>

---> quantity of water in 1 minute = 462 cubic inches

--->1 cubic feet = 1728 cubic inches

--->1 hour = 60 minutes,

<em>Step 2: Calculate feet per hour</em>

Quantity of water in 1 hour =  0.0347 x 462

---> quantity of water in 1 hour =  = 16.0314 cubic feet

---> Rounded off to--> 16 cubic feet

Therefore, 462 cubic inches per minute is 16 cubic feet of water per hour

5 0
1 year ago
Read 2 more answers
Use Definition 7.1.1, DEFINITION 7.1.1 Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = ∞ e−
poizon [28]

f(t)=\begin{cases}\cos t&\text{for }0\le t

The Laplace transform is then

\mathcal L_s\{f(t)\}=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=\int_0^\pi e^{-st}\cos t\,\mathrm dt

Let I denote the integral we want to compute. Integrating by parts, setting

u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt

\mathrm dv=\cos t\,\mathrm dt\implies v=\sin t

gives

\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\int_0^\pi e^{-st}\sin t\,\mathrm dt

Integrate by parts again, setting

u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt

\mathrm dv=\sin t\,\mathrm dt\implies v=-\cos t

Then

\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\left(-e^{-st}\cos t\bigg|_{t=0}^{t=\pi}-s\int_0^\pi e^{-st}\cos t\,\mathrm dt\right)

I=e^{-st}(\sin t-s\cos t)\bigg|_{t=0}^{t=\pi}-s^2I

(s^2+1)I=s(e^{-\pi s}+1)

I=\dfrac s{s^2+1}(e^{-\pi s}+1)

7 0
2 years ago
JL is a common tangent to circles M and K at point J. If angle MLK measures 61ᵒ, what is the length of radius MJ? Round to the n
Sladkaya [172]
3.122

The process to getting this comes in many steps. Firstly, you need to find the angles for JLK and MLJ. To find JLK use the arcsin function using the opposite side and the hypotenuse. 

Arcsin(Opp/Hype) = JLK
Arcsin(.5) = JLK
30 degrees = JLK

This means MLJ = 31 degrees since they add up to 61 degrees. 

Now we need to find the length of LJ, which we can do using the Pythagorean Theorem. 

3^2 + JL^2 = 6^2
9 + JL^2 = 36
JL^2 = 27
JL = \sqrt{27}

Now that we have the angle of MLJ and the length of JL, we can use the tangent function to find MJ.

Tan(angle) = opp/adj
Tan(31) = MJ/\sqrt{27}
\sqrt{27}Tan(31) = MJ
3.122 = MJ
7 0
2 years ago
Read 2 more answers
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