A street that is 40 yards long has a tree every 10 yards on both sides. How

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

igor_vitrenko [27]

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

$m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7$

So, the solution of the equation is

$y = Ae^{m_{1} t} + Be^{m_{2} t}$

where m₁ = 3 and m₂ = -7.

So,

$y = Ae^{3t} + Be^{-7t}$

Also,

$y' = 3Ae^{3t} - 7e^{-7t}$

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

6 0

3 years ago

Peter has built a gazebo, whose shape is a regular heptagon, with a side length of $1$ unit. He has also built a pathway around

Sholpan [36]

Answer:

Area of pathway = 10.375 unit²

Step-by-step explanation:

Given - Peter has built a gazebo, whose shape is a regular heptagon, with

a side length of 1 unit. He has also built a pathway around the

gazebo, of constant width 1 unit, as shown below.

To find - Find the area of the pathway.

Proof -

Central angle = $\frac{360}{7}$ = 51.43°

Now,

Central angle / 2 = $\frac{360}{14}$ = 25.71°

And

height = $\frac{1}{2}.\frac{1}{tan25.41}$ = 1.038

Now,

Inner area of gazebo = 7· $\frac{1}{2}$ ·1·(1.038) = 3.634 unit²

Outer area of gazebo = ( $\frac{2.038}{1.038}$ )²·(3.364) = 14.009 unit²

∴ we get

Area of pathway = Outer area - Inner area

= 14.009 - 3.634

= 10.375 unit²

⇒Area of pathway = 10.375 unit²

5 0

3 years ago

8x = 7x + 3<br><br> Would I combine the 8x and the 7x at all in this equation?

Kryger [21]

Yes...thats what u want to do. you want to get the x terms on one side and everything else on the other side.

8x = 7x + 3....so we will subtract 7x from both sides
8x - 7x = 7x - 7x + 3....simplify
1x, or just x = 3 (because ur 7x's cancelled each other out)

so ur answer is : x = 3

and u can check ur answer by subbing it back into the original equation.
8x = 7x + 3
8(3) = 7(3) + 3
24 = 21 + 3
24 = 24 (correct)

3 0

3 years ago

Read 2 more answers

Write the following in the figure.<br> 1. four hundredths<br> 2. two tenths

schepotkina [342]

What figure are you taking about,?

8 0

2 years ago

What is x squared equals 49/121

AleksAgata [21]

I think your answer would be x = 7/11, x = -7/11

7 0

3 years ago

Read 2 more answers