-6/16 + 1/4= <br> -3/8 -4/8<br> Can I have help with these questions?

Find the solution the each of the following first order linear differential equations:

Cerrena [4.2K]

Answer:

a. $y=\frac{2}{3}x^7+cx^4$

b. $y=2e^{7x}-ce^{5x}$

c. $y=e^{-2x}arctan(e^{2x})+ce^{-2x}$

d. $i=e^{-2t}\left(\frac{8\left(3e^{2t}\sin \left(3t\right)+2e^{2t}\cos \left(3t\right)\right)}{13}+C\right)$

Step-by-step explanation:

a) xy' -4y = 2 x^6

$xy'-4y=2x^6\\y'-\frac{4}{x}y=2x^5\\p(x)=\frac{-4}{x}\\Q(x)=2x^5\\\mu(x)=\int P(x)dx=\int \frac{-4}{x}dx=Ln|x|^{-4}\\y=e^{-\mu(x)}\int {e^{\mu(x)}Q(x)dx}\\y=x^4 \int {x^{-4}2x^6}dx\\y=\frac{2}{3}x^7+cx^4$

b) y' - 5y = 4e^7x

$y'-5y=4e^{7x}\\p(x)=-5\\Q(x)=4e^{7x}\\\mu(x)=\int P(x)dx=\int-5dx=-5x\\y=e^{-\mu(x)}\int {e^{\mu(x)}Q(x)dx}\\y=e^{5x}\int {e^{-5x}4e^{7x}}dx\\y=2e^{7x}-ce^{5x}$

c) dy/dx + 2y = 2/(1+e^4x)

$\frac{dy}{dx}+2y=\frac{2}{1+e^{4x}}\\p(x)=2\\Q(x)=\frac{2}{1+e^{4x}}\\\mu(x)=\int P(x)dx=\int 2dx=2x\\y=e^{-\mu(x)}\int {e^{\mu(x)}Q(x)dx}\\y=e^{-2x}\int {e^{2x}\frac{2}{1+e^{4x}}}dx\\y=e^{-2x}arctan(e^{2x})+ce^{-2x}$

d) 1/2 di/dt + i = 4cos(3t)

$\frac{1}{2}\frac{di}{dt}+i=4cos(3t)\\\frac{di}{dt}+2i=8cos(3t)\\p(t)=2\\Q(t)=8cos(3t)\\\mu(t)=\int P(t)dt=\int 2dt=2t\\i=e^{-\mu(t)}\int {e^{\mu(t)}Q(t)dt}\\i=e^{-2t}\int {e^{2t}8cos(3t}dt\\i=e^{-2t}\left(\frac{8\left(3e^{2t}\sin \left(3t\right)+2e^{2t}\cos \left(3t\right)\right)}{13}+C\right)$

3 0

3 years ago

The measure of an angle formed by intersecting chords is blank the sum of the measures of the intercepted arcs.

Mice21 [21]

Its the half, which is A

7 0

3 years ago

Read 2 more answers

What is the slope-intercept equation

viva [34]

The answer is y= -1/2 + (-4)

4 0

2 years ago

(3x-5)/(x-5)>=0 <br> How do I get the answers for this problem?

Diano4ka-milaya [45]

$\frac{3x-5}{x-5}$ > 0

First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like: $\frac{3x-5}{x-5}$ = 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x = $\frac{5}{3}$
Sixth, from the values of x above, we have these 3 intervals to test:
x < $\frac{5}{3}$
$\frac{5}{3}$ < x < 5
x > 5
Seventh, pick a test point for each interval.

1. For the interval x < $\frac{5}{3}$ :
Let's pick x - 0. Then, $\frac{3x0-5}{0-5}$ > 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.

2. For the interval $\frac{5}{3}$ < x < 5:
Let's pick x = 2. Then, $\frac{3x2-5}{2-5}$ > 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.

3. For the interval x > 5:
Let's pick x = 6. Then, $\frac{3x6-5}{6-5}$ > 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x < $\frac{5}{3}$ and x > 5

Answer: x < $\frac{5}{3}$ and x > 5

3 0

3 years ago

Solve 3x2−3x=0 by factoring.

o-na [289]

Answer:

3x(x-1) = 0

x= 1

Step-by-step explanation:

is 3x2 supposed to mean 3x^2 i dont know

5 0

3 years ago

-6/16 + 1/4= -3/8 -4/8 Can I have help with these questions?