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grandymaker [24]

3 years ago

12

Pls who knows how to do this? Make it correct just by adding one line5+5+5=550

1 answer:

Feliz [49]3 years ago

6 0

the answer shows in the picture

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Given the second order homogeneous constant coefficient equation y′′−4y′−12y=0 1) the characteristic polynomial ar2+br+c is r^2-

Vitek1552 [10]

Answer:

The initial value problem $y(x) = 4 e^{-2x} -3 e^{6 x}$

Step-by-step explanation:

<u>Step1:</u>-

a) Given second order homogenous constant co-efficient equation

$y^{ll} - 4y^{l}-12y=0$

Given equation in the operator form is $(D^{2} -4D-12)y=0$

<u>Step 2</u>:-

b) Let f(D) = $(D^{2} -4D-12)y$

Then the auxiliary equation is $(m^{2} -4m-12)=0$

Find the factors of the auxiliary equation is

$m^{2} -6m+2m-12=0$

m(m-6) + 2(m-6) =0

m+2 =0 and m-6=0

m=-2 and m=6

The roots are real and different

The general solution $y = c_{1} e^{-m_{1} x} + c_{2} e^{m_{2} x}$

the roots are $m_{1} = -2 and m_{2} = 6$

The general solution of given differential equation is

$y = c_{1} e^{-2x} + c_{2} e^{6 x}$

<u>Step 3</u>:-

C) Given initial conditions are y(0) =1 and y1 (0) =-26

The general solution of given differential equation is

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$ .....(1)

substitute x =0 and y(0) =1

$y(0) = c_{1} e^{0} + c_{2} e^{0}$

$1 = c_{1} + c_{2}$ .........(2)

Differentiating equation (1) with respective to 'x'

$y^l(x) = -2c_{1} e^{-2x} + 6c_{2} e^{6 x}$

substitute x= o and y1 (0) =-26

$-26 = -2c_{1} e^{0} + 6c_{2} e^{0}$

$-2c_{1} + 6c_{2} = -26$ .............(3)

solving (2) and (3) by using substitution method

$substitute c_{2} =1- c_{1}$ in equation (3)

$-2c_{1} + 6(1-c_{1}) = -26$

on simplification , we get

$-2c_{1} + 6(1)-6c_{1}) = -26$

$-8c_{1} = -32$

dividing by'8' we get $c_{1} =4$

substitute $c_{1} =4$ in equation $1 = c_{1} + c_{2}$

so $c_{2} = 1-4 = -3$

now substitute $c_{1} =4 and c_{2} =-3$ in general solution

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$

$y(x) = 4 e^{-2x} -3 e^{6 x}$

now the initial value problem

$y(x) = 4 e^{-2x} -3 e^{6 x}$

7 0

3 years ago

Can someone pls pls help me?

wlad13 [49]

Answer:

i cant see that lol

Step-by-step explanation:

ok

5 0

3 years ago

Fill in the blank with <,>,or,= to make the statement true -1 1/2 _ -1/4

Sladkaya [172]

Answer:

<

Step-by-step explanation:

Since it is a negative value, whatever fraction is closer to 0 will be greater.

7 0

3 years ago

Y=7×2^t/30 what is the equation solved for t

CaHeK987 [17]

Answer:

Option C

Step-by-step explanation:

Given expression in the question is,

y = $7(2)^{\frac{t}{30}}$

To get the value of t,

$\frac{y}{7}=2^{\frac{t}{30}}$

Take the log on both the sides of the expression,

log( $\frac{y}{7}$ ) = $\text{log}[(2^{\frac{t}{30}})]$

log( $\frac{y}{7}$ ) = $\frac{t}{30}(\text{log2})$

t = $30(\frac{\text{log}\frac{y}{7}}{\text{log}2} )$

t = $30[\text{log}_2(\frac{y}{7})]$ [Since, $\frac{\text{loga}}{\text{logb}}=\text{log}_a(b)}$ ]

Therefore, Option C will be the answer.

3 0

3 years ago

Please help!! lol y=?x+?

aev [14]

Answer:

y= 4x-9

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers