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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

Which triangle has a bigger area: 1. A triangle with sides measuring 300, 400, and 500. 2. A triangle with sides measuring 300,

Kitty [74]

A triangle with sides measuring 300, 400 and 700 because 300+400+700=1400 which is higher than the first triangle 300+400+500=1200

6 0

3 years ago

My Space Mathematician Teacher needs this answer,

GaryK [48]

219864070186 - 277409297400 = -57545227214 x 780498 = -4.4913935e+16 - 1274806769367926946610431 = -1.2748068e+24 x 3.6614422e+21 = -4.6676314e+45 - 437913 = -4.6676314e+45 - 88878 = -4.6676314e+45

Ф = 500 , 5,000

5,000 / ∞ x ∞ x ∞ = 5,000^+++∞/ 68307147 = 0.00007319878^+++∞ - 917 - 974 - 9 - 19344381 = -19346280.9999^+++∞

-4.6676314e+45 -19346280.9999^+++∞ = -4.6676314e+45^+++∞

I believe that will be your answer.

I tried my best.

4 0

3 years ago

What does this equal?

anzhelika [568]

It should be about 22.55.

4 0

3 years ago

Four friends purchase a pineapple for $2.89 and 18.4 pounds of peaches. The peaches cost $1.75 per pound. The friends share the

umka21 [38]

Answer: $3.00

Step-by-step explanation:

7 0

3 years ago

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