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3 years ago

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Plse help my grades go in tomorrow and I am currently struggling!!

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PilotLPTM [1.2K]3 years ago

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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

2 years ago

Which trigonometric function would you use to solve the problem?

umka21 [38]

you would use tan or cot

3 0

3 years ago

Which of the following cannot be used to express the number 8? eight more than zero +8 eight below zero positive eight

MrRa [10]

Eight below zero would express the value -8, negative eight,

but not 8.

7 0

2 years ago

Read 2 more answers

Which statement is true about the points and planes?

Reil [10]

Answer:

The answer is B

Step-by-step explanation:

Ed2020

3 0

2 years ago

If you flipped the graph y=x^2+2x-2 vertically, you would get the graph y=-(x^2 +2x-2). true or false

Mars2501 [29]

If you flipped the graph y=x^2+2x-2 vertically, you would get the graph y=-(x^2+2x-2) this is True.

5 0

2 years ago