Mathematics College

Find the solution of the given initial value problem:

(a) y' + 2y = te^{-2t}, y(1) = 0

(b) t^{3}y' + 4t^{2}y = e^{-t}, y(-1) = 0

Answers

Answer 1

Answer:

[tex](a)\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]

[tex](b)\ y(t)=\ (1-t)e^{-t}\ -\ 2e[/tex]

Step-by-step explanation:

(a) [tex]y'\ +\ 2y\ =\ te^{-2t},\ y(1)\ =\ 0[/tex]

[tex]=>\ (D+2)y\ =\ te^{-2t}[/tex]

To find the complementary function

D+2 = 0

=> D = -2

So, the complementary function can by given by

[tex]y_c(t)\ =\ C.e^{-2t}[/tex]

Now, to find particular integral

[tex](D+2)y_p(t)\ =\ te^{-2t}[/tex]

[tex]=>y_p(t)\ =\ \dfrac{ te^{-2t}}{D+2}[/tex]

[tex]=\ \dfrac{ te^{-2t}}{-2+2}[/tex]

= not defined

So,

[tex]y_p(t)\ =\ \dfrac{ t^2e^{-2t}}{D^2}[/tex]

[tex]=\ \dfrac{t^2e^{-2t}}{(-2)^2}[/tex]

[tex]=\ \dfrac{t^2e^{-2t}}{4}[/tex]

So, complete solution can be given by

[tex]y(t)\ =\ y_c(t)\ +\ y_p(t)[/tex]

[tex]=> y(t) =\ C.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]

As given in question

[tex]=>\ y(1)\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex]

[tex]=>\ 0\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex]

[tex]=>\ C\ =\ 4e^2[/tex]

Hence, the complete solution can be give by

[tex]=>\ y(t) =\ 4e^2.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]

[tex]=>\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]

(b) [tex]t^{3}y'\ +\ 4t^{2}y\ =\ e^{-t},\ y(-1)\ =\ 0[/tex]

[tex]=>\ y'\ +\ 4t^{-1}y\ =\ t^{-3}e^{-t}[/tex]

Integrating factor can be given by

[tex]I.F\ =\ e^{\int (4t^{-1})dt}[/tex]

[tex]=\ e^{log\ t^4}[/tex]

[tex]=\ t^4[/tex]

Now , the solution of the given differential equation can be given by

[tex]y(t)\times t^4\ =\ \int t^{-3}e^{-t}t^4dt\ +\ C[/tex]

[tex]=>\ y(t)\ =\ \int t.e^{-t}dt\ +\ C[/tex]

[tex]=\ (1-t)e^{-t}\ +\ C[/tex]

According to question

[tex]y(-1)\ =\ (1-(-1))e^1\ +\ C[/tex]

[tex]=>\ 0\ =\ 2e\ +\ C[/tex]

[tex]=>\ C\ =\ -2e[/tex]

Now, the complete solution of the given differential equation cab be given by

[tex]y(t)\ =\ (1-t)e^{-t}\ -\ 2e[/tex]

Answer 2

Answer:

a. [tex]y(t)=\frac{t^2e^{-2t}}{2}-\frac{1}{2}e^{-2t}[/tex]

b.[tex]y=-t^{-3}e^{-t}-t^{-4}e^{-t}[/tex]

Step-by-step explanation:

We are given that

a.[tex]y'+2y=te^{-2t},y(1)=0[/tex]

Compare with [tex]y'+P(t)y=Q(t)[/tex]

We have P(t)=2,Q(t)=[tex]te^{-2t}[/tex]

Integration factor=[tex]\int e^{2dt}=e^{2t}[/tex]

[tex]y\cdot I.F=\int Q(t)\cdot I.F dt+C[/tex]

Substitute the values then, we get

[tex]y\cdot e^{2t}=\int te^{-2t}\cdot e^{2t} dt+C[/tex]

[tex]y\cdot e^{2t}=\int tdt+C[/tex]

[tex]ye^{2t}=\frac{t^2}{2}+C[/tex]

Substitute the values x=1 and y=0

Then, we get [tex]0\cdot e^2=\frac{1}{2}+C[/tex]

[tex]C=-\frac{1}{2}[/tex]

Substitute the value in the given function

[tex]ye^{2t}=\frac{t^2}{2}-\frac{1}{2}[/tex]

[tex]y=\frac{t^2}{2}e^{-2t}-\frac{1}{2}e^{-2t}[/tex]

Hence, [tex]y(t)=\frac{t^2e^{-2t}}{2}-\frac{1}{2}e^{-2t}[/tex]

b.[tex]t^3y'+4t^2y=e^{-t},y(-1)=0[/tex]

[tex]y'+\frac{4}{t}y=\frac{e^{-t}}{t^3}[/tex]

[tex]P(t)=\frac{4}{t},Q(t)=\frac{e^{-t}}{t^3}[/tex]

I.F=[tex]\int e^{\frac{4}{t}dt}=e^{4lnt}=e^{lnt^4}=t^4[/tex]

[tex]y\cdot \frac{t^4}=\int e^{-t}\frac{t^4}{t^3} dt+C[/tex]

[tex]y\cdot t^4=\int te^{-t}dt+C[/tex]

[tex]yt^4=-te^{-t}+\int e^{-t} dt+C[/tex]

[tex]u\cdot v dt=u\int vdt-\int (\frac{du}{dt}\cdot \int vdt)dt[/tex]

[tex]yt^4=-te^{-t}-e^{-t}+C[/tex]

Substitute the values x=-1,y=0 then, we get

[tex]0=-(-1)e-e+C[/tex]

[tex]C+e-e=0[/tex]

C=0

Substitute the value of C then we get

[tex]yt^4=-te^{-t}-e^{-t}[/tex]

[tex]y=-t^{-3}e^{-t}-t^{-4}e^{-t}[/tex]

Related Questions

A random sample of 50 consumers taste tested a new snack food. Their responses were coded (0: do not like; 1: like, 2: indifferent) and recorded below: a. Test H0: p = 0.5 against Ha: p > 0.5, where p is the proportion of customers who do not like the snack food (n=17). Use α = 0.10. b. Find the observed significance level of your test.

Answers

Answer:

The level of significance observed is 0.99154

Step-by-step explanation:

Assuming that in a sample of size 50 people stated that they do not like the snack (p = 17/50), you have:

Proportion in the null hypothesis [tex]\pi_0=0.5[/tex]

Sample size [tex]n=50[/tex]

Sample proportion [tex]p=17/50=0.34[/tex]

The expression for the calculated statistic is:

[tex] = \frac{(p - \pi_0)\sqrt{n}}{\sqrt{\pi_0(1-\pi_0)}}[/tex]

[tex]= \frac{(0.34 - 0.5)\sqrt{50}}{\sqrt{0.34(0.66)}} = -2,38833[/tex]

The level of significance observed is obtained from the value of the statistic calculated:

[tex]P(Z>Z_{calculated}) = 0.99154[/tex]

In a sports shop there are T-shirts of 5 different colors, shorts of 4 different colors, and socks of 3 different colors. How many different uniforms can you compose from these items?

Answers

Answer: 60

Step-by-step explanation:

5 shorts × 4 shorts × 3 socks = 60 different uniforms

Convert into decimals (use your calculator): a) 6/38 b) 45/55 c) 4/28 d) 35/28 e) 1030/2030

Answers

Answer:

[tex]a.\hspace{3} \frac{6}{38} = 0.15789474\\\\b.\hspace{3} \frac{45}{55} = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 0.14285714\\\\d.\hspace{3} \frac{35}{28} = 1.25\\\\e.\hspace{3} \frac{1030}{2030} = 0.50738916[/tex]

Step-by-step explanation:

Rational numbers, expressed as decimal numbers, are obtained from the operation of division between the integer of the numerator and the integer of the denominator. Then:

[tex]a.\hspace{3} \frac{6}{38} = 6\div38 = 0.15789474\\\\ b.\hspace{3} \frac{45}{55} = 45\div55 = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 4\div28 = 0.14285714\\\\ d.\hspace{3} \frac{35}{28} = 35\div28 = 1.25\\\\ e.\hspace{3} \frac{1030}{2030} = 1030\div2030 = 0.50738916[/tex]

UBER® is an American multinational online transportation network company. The fare for any trip and the different services that the company offers is calculated by adding base fare, time and distance. However, in certain cities, the company has a service called UberRUSH and it charges the client based on a base fare and a charge per mile. During daytime in New York City, UberRUSH has a base fare of $3.00plus $2.50per mile. Sergio just arrived into John F. Kennedy International Airport and he wants to determine a function for his transportation costs during the day using UberRUSH.

Part A: Write a linear function to describe the cost of using UberRUSH services during the day in New York City.

Part B: What is the rate of change for the cost of riding UberRUSH during the day in New York City and what does the rate of change represent?

Part C: What does the ࠵?-intercept represent in the cost function you wrote?

Part D:Represent the situation on the graph below.

Answers

Answer:

See explanation

Step-by-step explanation:

During daytime in New York City, UberRUSH has a base fare of $3.00 plus $2.50 per mile.

A. Let x be the number of miles Sergio uses the taxi, then he will pay $2.50x for x miles plus a base fare of $3.00. In total, the cost of using UberRUSH services during the day in New York City is

[tex]C=3.00+2.50x[/tex]

B. The rate of change for the cost of riding UberRUSH during the day in New York City is the slope of the line represented by linear function in part A. So, the rate of change is 2.5 and it shows the change in price per 1 mile driven.

C. The y-intercept is 3.00 (or simply 3) and it represents the initial cost (when 0 miles are driven)

D. The graph of the function is shown in attached diagram. The diagram shows the part of the line starting from point (0,3). This is because the number of miles driven cannot be negative.

Write the following as a ratio: $72 for 488 photos

Answers

Answer:

The required ratio is 9 : 61

Step-by-step explanation:

Given,

$72 for 488 photos,

That is, the price of 488 photos = 72 dollars,

So, the ratio of price of photos and number of photos = [tex]\frac{72}{488}[/tex]

∵ HCF(72, 488) = 8,

Thus, the ratio of price of photos and number of photos = [tex]\frac{72\div 8}{488\div 8}[/tex]

= [tex]\frac{9}{61}[/tex]

PLEASE HELP ME I WILL GET SENT HOME IF I DONT PASS

Answers

Answer: p = -0.004n + 8

Step-by-step explanation:

Consider the number of shirts & the price as an x,y - coordinate

4000 shirts at $32 --> (4000, 32)

5000 shirts at $28 --> (5000, 28)

Use the slope formula: [tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

[tex]m=\dfrac{28-32}{5000-4000}\quad =\dfrac{-4}{1000}\quad =-0.004[/tex]

Next, use the Point-Slope formula (replace x with n & replace y with p):

[tex]y-y_1=m(x-x_1)\implies p-p_1=m(n-n_1)\\\\\\p-28=-0.004(n-5000)\\\\p-28=-.004n-20\\\\\large\boxed{p=-0.004n+8}[/tex]

A prescription calls for the following: Sodium citrate 5 g Tartar Emetic 0.015 g Cherry syrup ad 120 mL Using a balance with a sensitivity of 4 mg, an acceptable weighing error of 5% and cherry syrup as the solvent for tartar emetic, how could you obtain the correct quantity of tartar emetic to fill the prescription?

Answers

Final answer:

To measure tartar emetic for the prescription, keep in mind the sensitivity of your balance and the acceptable weighing error. Tartar emetic of 15mg should be weighed as accurately as the balance allows and if the measurement falls within the acceptable error range, it can be dissolved in cherry syrup for uniform distribution.

Explanation:

The tartar emetic dose required by the prescription is 0.015 g. However, you're using a balance with a sensitivity of 4 mg (milligrams). Similar to procure the accurate tartar emetic measurement, you must take into account the balance's sensitivity and the acceptable weighing error of 5%.

0.015 g is equal to 15 mg. Our balance's lowest measurement is 4 mg, which is less than the 15 mg we need to weight accurately. When considering the acceptable weighing error of 5%, the actual amount of tartar emetic measured could range from 14.25 mg to 15.75 mg (15 mg ± 5%).

To solve with the stipulated constraints, you need to carefully weigh out the tartar emetic as close as the balance will allow, and if it does not reach the desired measurement but falls within the acceptable error range, you can dissolve it in the cherry syrup for uniform distribution within the medication. Always remember to mix well and be precise with measurements to ensure the correct strength of medicine.

Learn more about Medicine Preparation here:

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Final answer:

Calculate the acceptable error range for the weight of tartar emetic, then weigh it out on a balance with the specified sensitivity. Once within the error range, dissolve the tartar emetic in the cherry syrup.

Explanation:

The question asks how to obtain the correct quantity of tartar emetic given the equipment and accepted weighing error. To do this, first you must calculate the acceptable error range by multiplying the desired weight of the tartar emetic (0.015 g) by 5%, which yields an acceptable error of 0.00075 g. Since your balance has a sensitivity of 4 mg (0.004 g), it can accurately measure out this quantity. Then you would weigh out the tartar emetic on the balance until you are within this error range.

It's important to note that tartar emetic is fully soluble in water and so cherry syrup, which is water-based, can be used as the solvent for tartar emetic. After weighing, dissolve the tartar emetic into the cherry syrup before moving on to measure out the other ingredients for the prescription.

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35 year old female. 5’8', 220 pounds. Calculate Adjusted Body Weight. (Round to the nearest tenth if applicable): ________

Answers

Answer:

Her adjusted body weight is 78.26kg = 78.3kg, rounded to the nearest tenth.

Step-by-step explanation:

To find the Adjusted Body Weight(AjBW), first we have to consider the Ideal Body Weight(IBW).

We have the following formulas:

[tex]IBW = 45.5kg + 2.3kg*i[/tex], in which i is the number of inches that the woman has above 5 feet.

[tex]AjBW = IBW + 0.4*(ABW - IBW)[/tex], in which [tex]ABW[/tex] is her Actual Body Weight

So, in this problem:

Each pound has 0.45kg. So her weight is [tex]ABW = 220*0.45 = 99.8[/tex]kg.

The woman is 8 inches above 5 feet, so [tex]i = 8[/tex].

[tex]IBW = 45.5 + 2.3*8[/tex]

[tex]IBW = 63.9[/tex]kg

Her ideal body weight is 63.9kg. Her adjusted body weight is:

[tex]AjBW = IBW + 0.4*(ABW - IBW)[/tex]

[tex]AjBW = 63.9 + 0.4*(99.8-63.9)[/tex]

[tex]AjBw = 78.26[/tex]kg

Her adjusted body weight is 78.26kg = 78.3kg.

EXAMPLE 2 The arc of the parabola y = 3x2 from (5, 75) to (10, 300) is rotated about the y-axis. Find the area of the resulting surface. SOLUTION 1 Using y = 3x2 and dy dx = we have, from this formula, S = 2πx ds = 10 2πx 1 + dy dx 2 dx 5 = 2π 10 x 1 + 36x2 dx 5 . Substituting u = 1 + 36x2, we have du = dx. Remembering to change the limits of integration, we have S = π 36 3601 u du 901 = π 36 3601 901 = . SOLUTION 2 Using x = y 3 and dx dy = we have S = 2πx ds = 300 2πx 1 + dx dy 2 dy 75 = 2π 300 y 3 1 + 1 12y dy 75 = π 3 300 12y + 1 dy 75 = π 36 3601 u du 901 (where u = 1 + 12y) = (as in Solution 1)

Answers

Final answer:

To find the surface area of the resulting surface when rotating the arc of a parabola about the y-axis, you can integrate using either x or y as the variable. Both methods yield the same result.

Explanation:

The question asks for the area of the surface created by rotating the arc of the parabola y = 3x^2 from (5, 75) to (10, 300) about the y-axis.

There are two different solutions provided, both utilizing different methods of integration.

The first solution uses the given equation y = 3x^2 and integrates with respect to x, while the second solution uses the equation x = y^(1/3) and integrates with respect to y.

Both solutions arrive at the same answer to find the surface area.

Let's use the same bag of 12 balls: S are red, 3 are brown, 2 are green and 2 are white. If you draw a red ball and put it aside: a) How many balls are left in the bag? b) How many of the balls in the bag are red? c) How many are brown? Green? White? d) Draw a second ball from the bag. What is the probability that it is green? (Hint: How many of the balls in the bag are green? How many balls are in the bag?) Now put both balls back in the bag and draw two balls without replacing them. e) What is the probability that the first ball is red and the second ball is green (without replacement)? What is the probability that the first ball is red and the second ball is also red (without replacement)? f) What is the probability that the first ball is green and the second ball is also green (without replacement)? g) What is the probability that the first ball is brown and the second ball is white (without replacement)? Would this be different than the probability that the first ball is white and the second ball is brown (without replacement)? Explain. (Feel free to use calculations in your explanation.) h) 27

Answers

Answer:

a) There are 11 balls left in the bag.

b) 4 of the balls in the bag are red.

c) 3 brown, 2 green, 2 white.

d) [tex]P = \frac{2}{11}[/tex]

e) The probability that the first ball is red and the second is green is:

[tex]\frac{5}{66}[/tex]

The probability that both balls are red is:

[tex]\frac{5}{33}[/tex]

f) The probability that both balls are green is:

[tex]\frac{1}{66}[/tex]

g) The probability that the first ball is brown and the second is white is:

[tex]\frac{1}{22}[/tex]

The probability that the first ball is brown and the second is white is:

[tex]\frac{1}{22}[/tex]

They are the same probabilities.

Step-by-step explanation:

There are 12 balls in the bag

5 red

3 brown

2 green

2 white

If you draw a red ball and put it aside:

a) How many balls are left in the bag?

There were 12 balls in the bag, and one was put aside.

So, there are 11 balls left in the bag.

b) How many of the balls in the bag are red?

There were 5 red balls in the bag, and one was put aside.

So, there are 4 red balls in the bag.

c) How many are brown? Green? White?

The ball put aside was red, so we still have the same number of the balls of the other colors.

3 brown, 2 green, 2 white.

d) Draw a second ball from the bag. What is the probability that it is green?

There are 11 balls in the bag, 2 of which are green. So the probability that the second ball is green is:

[tex]P = \frac{2}{11}[/tex]

Now put both balls back in the bag and draw two balls without replacing them.

e) What is the probability that the first ball is red and the second ball is green (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is red. There are 12 balls, 5 of which are red. So:

[tex]P_{1} = \frac{5}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 2 of which are green. So:

[tex]P_{2} = \frac{2}{11}[/tex]

The probability that the first ball is red and the second is green is:

[tex]P = P_{1}*P_{2} = \frac{5}{12}*\frac{2}{11} = \frac{5}{66}[/tex]

What is the probability that the first ball is red and the second ball is also red (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is red. There are 12 balls, 5 of which are red. So:

[tex]P_{1} = \frac{5}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 4 of which are red. So:

[tex]P_{2} = \frac{4}{11}[/tex]

The probability that both balls are red is:

[tex]P = P_{1}*P_{2} = \frac{5}{12}*\frac{4}{11} = \frac{5}{33}[/tex]

f) What is the probability that the first ball is green and the second ball is also green (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is green. There are 12 balls, 2 of which are green. So:

[tex]P_{1} = \frac{2}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 1 of which is green. So:

[tex]P_{2} = \frac{1}{11}[/tex]

The probability that both balls are green is:

[tex]P = P_{1}*P_{2} = \frac{2}{12}*\frac{1}{11} = \frac{1}{66}[/tex]

g) What is the probability that the first ball is brown and the second ball is white (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is brown. There are 12 balls, 3 of which are brown. So:

[tex]P_{1} = \frac{3}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 2 of which are white. So

[tex]P_{2} = \frac{2}{11}[/tex]

The probability that the first ball is brown and the second is white is:

[tex]P = P_{1}*P_{2} = \frac{3}{12}*\frac{2}{11} = \frac{1}{22}[/tex]

Would this be different than the probability that the first ball is white and the second ball is brown?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is white. There are 12 balls, 2 of which are brown. So:

[tex]P_{1} = \frac{2}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 3 of which are brown. So

[tex]P_{2} = \frac{3}{11}[/tex]

The probability that the first ball is brown and the second is white is:

[tex]P = P_{1}*P_{2} = \frac{2}{12}*\frac{3}{11} = \frac{1}{22}[/tex]

They are the same probabilities.

At a restaurant a menu has 5 salads and 6 entrees. How many ways can you order a dinner that contains 1 salad and 1 entree?

Answers

Answer: 30

Step-by-step explanation:

The combination of n things taking r at a time is given by :-

[tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]

Given : At a restaurant a menu has 5 salads and 6 entrees.

Then, the number of ways you can order a dinner that contains 1 salad and 1 entree will be :_

[tex]^5C_1\times ^6C_1\\\\=5\times6=30[/tex]

Hence, the number of ways you can order a dinner that contains 1 salad and 1 entree = 30

Consider a particle moving around a circle with a radius of 38cm. It rotates from 10 degrees to 100 degrees in 11 seconds. Calculate the instantaneous velocity of the particle.

Answers

Step-by-step explanation:

Given that,

Radius of circle, r = 38 cm = 0.38 m

It rotates form 10 degrees to 100 degrees in 11 seconds i.e.

[tex]\theta_i=10^{\circ}=0.174\ rad[/tex]

[tex]\theta_f=100^{\circ}=1.74\ rad[/tex]

Let [tex]\omega[/tex] is the angular velocity of the particle such that, [tex]\omega=\dfrac{\omega_f-\omega_i}{t}[/tex]

[tex]\omega=\dfrac{1.74-0.174}{11}[/tex]

[tex]\omega=0.142\ rad/s[/tex]

We need to find the instantaneous velocity of the particle. The relation between the angular velocity and the linear velocity is given by :

[tex]v=r\times \omega[/tex]

[tex]v=0.38\times 0.142[/tex]

v = 0.053 m/s

So, the instantaneous velocity of the particle is 0.053 m/s. Hence, this is the required solution.

The measure of the complement of an angle is 14 less than half the measure of its supplement. Find the angle.

Answers

Answer:

The angle is 28.

Step-by-step explanation:

If we consider the system below (were "A"is the angle):

A + [(x/2)-14] = 90 (complement)

A + x = 180 (suplement)

We can isolate "X=180 - A" and apply this in the first equation. This way we are able to have only one variable, and w'll see that A=28.

If the pressure inside a rubber balloon is 1500 mmHg, what is this pressure in pounds-force per square inch (psi)? Answer: 29.0 psi

Answers

Answer:

Step-by-step explanation:

You already have the answer.

1 [at] = 760 [mmHg]

1 [at] = 14,7 [psi]

1500 [mmHg] * 1 [at] / 760 [mmHg] * 14,7 [psi] / 1 [at] = 29.0 [psi]

Verify that the function(s) solve the following differential equations (DES): a) y' = -5y; y = 3e-5x b) y' = cos(3x); y = į sin(3x) + 7 c) y' = 2y; y = ce2x , where c is any real number. d) y" + y' – 6y = 0 ; yı = (2x, y2 = (–3x e) y" + 16y = 0; yı = cos(4x), y2 = sin(4x)

Answers

Answer:

In the step-by-step explanation, the verifications are made.

Step-by-step explanation:

a) [tex]y' = -5y[/tex]

This one can be solved by the variable separation method

[tex]y' = -5y[/tex]

[tex]\frac{dy}{dx} = -5y[/tex]

[tex]\frac{dy}{y} = -5dx[/tex]

[tex]\int \frac{dy}{y} = \int {-5} \, dx[/tex]

[tex]ln y = -5x + C[/tex]

[tex]e^{ln y} = e^{-5x + C}[/tex]

[tex]y = Ce^{-5x}[/tex]

The value of C is the value of y when x = 0. If [tex]y(0) = 3[/tex], then we have the following solution:

[tex]y = 3e^{-5x}[/tex]

b) [tex]y' = cos(3x)[/tex]

This one can also be solved by the variable separation method

[tex]y' = cos(3x)[/tex]

[tex]\int y' \,dy = \int {cos(3x)} \, dx[/tex]

[tex]y = \frac{sin(3x)}{3} + K[/tex]

K is also the value of y, when x = 0. So, if [tex]y(0) = 7[/tex], we have the following solution.

[tex]y = \frac{sin(3x)}{3} + 7[/tex]

c) [tex]y' = 2y[/tex]

Another one that can be solved by the variable separation method

[tex]y' = 2y[/tex]

[tex]\frac{dy}{dx} = 2y[/tex]

[tex]\frac{dy}{y} = 2dx[/tex]

[tex]\int \frac{dy}{y} = \int {2} \, dx[/tex]

[tex]ln y = 2x + C[/tex]

[tex]e^{ln y} = e^{2x + C}[/tex]

[tex]y = Ce^{2x}[/tex]

C is any real number depending on the initial conditions.

d) [tex]y'' + y' - 6y = 0[/tex]

Here, the solution depends on the roots of the following equation:

[tex]r^{2} + r - 6 = 0[/tex]

[tex]r = \frac{-1 \pm 5}{2}[/tex]

[tex]r = -3[/tex] or [tex]r = 2[/tex].

So the solution is

[tex]y(t) = c_{1}e^{-3t} + c2e^{2t}[/tex]

The values of [tex]c_{1}, c_{2}[/tex] depends on the initial conditions.

e) [tex]y'' + 16y = 0[/tex]

Again, we find the roots of the following equation:

[tex]r^{2} + 16 = 0[/tex]

[tex]r^{2} = -16[/tex]

[tex]r = \pm 4i[/tex]

So we have the following solution

[tex]y(t) = c_{1}cos(4t) + c_{2}sin(4t)[/tex]

The values of [tex]c_{1}, c_{2}[/tex] depends on the initial conditions.

A) Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is needed to make your meaning clear. For example, 1⋅105 is not the same as 1105. When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol ×. For example, for the expression ma, typing m ⋅ a would be correct, but mxawould be incorrect.

ma=....

B) Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

2cos2(θ)−1 =.....

c) Enter the expression asin(x)+15, where asin(x) is the inverse sine function.

asin(x)+15 =....

D) Enter the expression (√(2gΔy)/m), where Δ is the uppercase Greek letter Delta.

(√2gΔy/m)=........

E) Enter the expression N0e−λt, where N0 is N-naught (an N with a subscript zero) and λ is the lowercase Greek letter lambda.

Answers

The correct representation are as follows:

Part (a): [tex]\boxed{m\cdot a=}[/tex]

Part (b): [tex]\boxed{2*\left(\text{cos}(2\cdot \theta)\right)-1=}[/tex]

Part (c): [tex]\boxed{a*\left(\text{sin}(x)\right)+15=}[/tex]

Part (d): [tex]\boxed{\sqrt{\dfrac{2\cdot g\cdot \triangle y}{m}}=}[/tex]

Part (e): [tex]\boxed{N_{0}\cdot e^{-(\lambda\cdot t)}}[/tex]

Further explanation:

in the question it is given that while writing any mathematical expression the symbol to be used for multiplication is [tex]\cdot\text{ dot}[/tex].

For example: [tex](2\times 3)[/tex] is incorrect and [tex](2\cdot 3)[/tex] is correct.

If a mathematical expression is to be written in its explicit form then the symbol [tex]*\text{ aestrick}[/tex] is used.

Part (a):

The expression given in part (a) is as follows:

[tex]ma=[/tex]