What is the value of x? 3 6 27 72

Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

3 years ago

Add 5+[-8] what is the awenser or sum.

Arte-miy333 [17]

Answer:

Your answer is -3

Step-by-step explanation:

We know your answer is going to be the answer because 8 is a bigger number than 5, so you take 5 away from (-8)

Which leaves you with -3

Hope this helps!

-Payshence

7 0

4 years ago

Which value is in the solution set of |x|+9=-2

Eduardwww [97]

Answer:

The equation has no solutions

Step-by-step explanation:

If we re arrange the equation a little more we will obtain the following

|x|=-11

Which is a contradiction since, the |x| is always a positive number

Attached is the representation of this function y= |x|

It means, that it doesnt matter what value 'x' has, the answer is the same number but, positive

4 0

4 years ago

Can someone help me? :(<br><br>

aleksandr82 [10.1K]

Answer:

0.037037037037037 i think

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Help with geometry! ! !! ! !

snow_lady [41]

Answer:

RST

Step-by-step explanation:

Look at the angles for M, N and O. They are equal to 79, 60, and 42 degrees, respectively. Now look at the triangle to the right of triangle MNO. In what order should we line up the letters so that the angles for that triangle should also equal 79, 60 and 42 degrees? If you look at it for a second, you should that this problem is easy to solve! Did you get it right! That's right, by lining up the angles in the order RST, the measure of each angle is equal to the measures of MNO.

Give me Brainliest and have a great day! :)

Cheers!

8 0

4 years ago