Question 3 Unsaved

Mathematics

1 answer:

bazaltina [42]2 years ago

8 0

Answer:

Step-by-step explanation:

DE given is

y''+8y'+15y=0, y(0)=0, y'(0)=1

Take Laplace on the DE

We get

$s^2 Y(s) -sY(0) -y'(0)+8(sY(s)-y(0))+15Y(s) =0\\s^2 Y(s) -s(0) -1+8(sY(s)-0)+15Y(s) =0\\Y(s)(s^2+8s+15)-1=0\\Y(s) = \frac{1}{s^2+8s+15}$

Simplify to get

Y(s) = $\frac{1}{2}[ {\frac{1}{s+3} -\frac{1}{s+5} }]$

Take inverse

$y(t) = \frac{e^{-3t} -e^{-5t}}{2}$

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Suppose y = 2x + 1 , where x and y are functions of t. (a) if dx/dt = 3, find dy/dt when x = 4. dy dt = (b) if dy/dt = 2, find d

qwelly [4]

Answer:

Step-by-step explanation:

If y = 2x + 1, then dy/dt = 2(dx/dt).

If y = 2x + 1, then y = 2(40) + 1 when 40 is substituted for x. y = 81.

(a) if dx/dt = 3, find dy/dt when x = 4:

Replacing dx/dt with 3 in dy/dt = 2(dx/dt) yields dy/dt = 2(3) = 6.

(b) if dy/dt = 2, find dx/dt when x = 40:

Replacing dy/dt with 2 in dy/dt = 2(dx/dt) results in 2 = 2(dx/dt), so dx/dt must be 1.

3 0

2 years ago

Find the reference angle for the given angle. Show your work. -404°

Mamont248 [21]

Answer:

44°

Step-by-step explanation:

The reference angle is the magnitude of the smallest angle between the terminal ray and the x-axis.

Your angle is one full circle clockwise plus 44° more (-360° -44° = -404°), so makes an angle of -44° with respect to the positive x-axis. The magnitude of that is 44°.

_____

Using a function

The attachment shows a calculator function that can be used to find the reference angle for any angle. (Replace the value -404 with the angle of your choice.)