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3 years ago

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When a trigonometric equation is solvable, it often has infinitely many solutions; and (depending on the context) the first solu

tion we find may not be the one we want. Explain how, given one solution of a trigonometric equation of a trigonometric equation (say, one that reduces down to ), we can find other solutions from the first solution we find, without resorting to a calculator.'

1 answer:

Anit [1.1K]3 years ago

3 0

Answer:

(See explanation for further details)

Step-by-step explanation:

An approach is handling the trigonometric equation so that number of trigonometric functions can be reduce to one, in order to determine the periodicity of complete expression and therefore, to determine the complete set of solutions.

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Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago

SOMEONE HELP BECAUSE I DONT GET THIS

Leno4ka [110]

Answer:

x = 18

m<A = 166°

Step-by-step explanation:

The parallel lines are cut by a transversal, so corresponding angles A and B are congruent.

7x + 40 = 3x + 112

4x = 72

x = 18

m<A = 7x + 40 = 7(18) + 40 = 126 + 40 = 166

6 0

2 years ago

Read 2 more answers

It’s not supplementary does anyone know what it is?

Mrac [35]

I agree with your answer "supplementary". The two angles combine to form 180 degrees, aka a straight angle

Your teacher seemed to have made a typo.

3 0

3 years ago

A piece of string 25 1/2 inches long will be cut into 3/4 pieces. How many pieces will there be?

FromTheMoon [43]

25 1/2 is 51/2, or 102/4. If we're cutting into 3/4 inch pieces, we have (102/4)/(3/4)=34 pieces.

5 0

3 years ago

Can someone help me with this please

Dmitriy789 [7]

for 7

Yes it is right cuz 16 is the double of 8 and even the price if doubled up so the price is right

for 8

Apply the formula

$2\sqrt{\frac{Area}{\pi } }$

$2\sqrt{\frac{154}{\pi } }$

$2\sqrt{49.02}$

$2(7)$

$14$

the diameter is 14

Hope this helps :)

7 0

2 years ago