What is the value for Z in the solution of the system?

Solve the following differential equation: y" + y' = 8x^2

PtichkaEL [24]

Answer:

$y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$

Step-by-step explanation:

Let's find a particular solution. We need a function of the form $y= ax^3+bx^2+cx+d$ such that

$y'= 3ax^2+2bx+c$ and

$y''= 6ax+2b$

$y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2$

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a = 8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is $y_p= \frac{8}{3}x^3-8x^2+16x$ .

Now, let's find the solution $y_p$ of the homogeneus equation $y''+y'=0$ with the method of constants coefficients. Let $y=e^{\lambda x}$

$y'=\lambda e^{\lambda x}$

$y''=\lambda^2e^{\lambda x}$

then $\lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0$

$e^{\lambda x}(\lambda +\lambda^2)= 0$

$(\lambda +\lambda^2)= 0$

$\lambda (1+\lambda)= 0$

$\lambda =0$ and $\lambda)= -1$ .

So, $y_h = C_1 + C_2e^{-x}$ and the solution is

$y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$ .

5 0

3 years ago

Finding Slope On a coordinate plane, a line goes through points (0, 1) and (4, 2). What is the slope of the line? m =

ivolga24 [154]

Answer:

slope = $\frac{1}{4}$

Step-by-step explanation:

Calculate the slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (0, 1) and (x₂, y₂ ) = (4, 2)

m = $\frac{2-1}{4-0}$ = $\frac{1}{4}$

7 0

3 years ago

What do two equivalent expression always have in common

Andre45 [30]

<span> according to the code of Mathematical Logic, there exists more properties, one of its principles is the consequence and effects property, or the so-called equivalence implicationwhich is defined as follows:Two expressions A and B are equivalent if A can produce B, and B can produce A. It means A implies B and B implies A.</span>

6 0

3 years ago

5y+33+6y into distributive property

Elis [28]

Answer : 44y. Put why because it’s repeated

7 0

3 years ago

Read 2 more answers

In a study relating the degree of warping, in mm, of a copper plate (y) to temperature in °C (x), the following summary statisti

elena-14-01-66 [18.8K]

Answer:

a)

y = 0.3035 + 0.0082x

b)

0.6315 mm

c)

x = 23.9634 °C

Step-by-step explanation:

a. Compute the least-squares line for predicting warping from temperature. Round the answers to four decimal places.

We need to find an equation of the form

y = b + mx

where m is the slope and b the Y-intercept.

The slope m can be computed with the formula

$\bf m=\displaystyle\frac{\displaystyle\sum_{i=1}^n(x_i-\bar x)(y_i-\bar y)}{\displaystyle\sum_{i=1}^n(x_i-\bar x)^2}$

Replacing the values in our formula (we will round at the end of the calculations)

$\bf m=\displaystyle\frac{806.94}{98,775}=0.008169476$

the Y-intercept b is computed with the formula

$\bf b=\bar y-m\bar x$

therefore we have

$\bf b=0.5188-0.008169476*26.36=0.303452613$

and the least-squares line rounded to 4 decimals would be

y = 0.3035 + 0.0082x

b. Predict the warping at a temperature of 40°C. Round the answer to three decimal places.

We simply replace x with 40 to get

y = 0.3035 + 0.0082*40 = 0.6315 mm

c. At what temperature will we predict the warping to be 0.5 mm? Round the answer to two decimal places

Here, we replace y with 0.5 and solve for x

0.5 = 0.3035 + 0.0082x ===> x = (0.5-0.3035)/0.0082 ===>

x = 23.9634 °C

6 0

3 years ago