Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Can this differential equation be solved using separation of variables?

(dy)/(dx)=2sin(x)-3cos(y)
Choose 1 answer:
(A) Yes
(B) No

Can this differential equation be solved using separation of variables? $\newline$ $\frac{d y}{d x}=2 \sin (x)-3 \cos (y)$ $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) No

View step-by-step help

View step-by-step help

Intermediate Value Theorem

Full solution

Q. Can this differential equation be solved using separation of variables?

\newline

\frac{d y}{d x}=2 \sin (x)-3 \cos (y)

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Check Equation Form: To determine if the given differential equation can be solved using separation of variables, we need to see if we can express the equation in the form of a product of a function of $x$ and a function of $y$ , such that we can integrate both sides separately. Let's try to rearrange the equation.
Separate Variables Attempt: We start by trying to separate the variables $y$ and $x$ on different sides of the equation. We want to get all the $y$ terms on one side and all the $x$ terms on the other side. We can attempt to do this by dividing both sides by $-3\cos(y)$ and multiplying by $dx$ to get $dy$ on one side.
Variable Separation Failure: After attempting to separate the variables, we would have: $\newline$ $\frac{dy}{-3\cos(y)} = \frac{2\sin(x)}{-3\cos(y)}dx$ $\newline$ However, we notice that the right side of the equation still contains both $x$ and $y$ , which means we cannot separate the variables as required for the method of separation of variables.
Conclusion: Since we cannot express the equation in a form where each side contains only one variable and its differentials, the differential equation cannot be solved using separation of variables.

More problems from Intermediate Value Theorem

Question

f

be a continuous function on the closed interval

[1,5]

f(1)=1

f(5)=-3

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

f(c)=2

for at least one

c

-3

1

\newline

f(c)=-2

for at least one

c

-3

1

\newline

f(c)=2

for at least one

c

1

5

\newline

f(c)=-2

for at least one

c

1

5

Posted 3 months ago

Question

h

be a continuous function on the closed interval

[-3,4]

h(-3)=-1

h(4)=2

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

h(c)=1

for at least one

c

-3

4

\newline

h(c)=-2

for at least one

c

-3

4

\newline

h(c)=-2

for at least one

c

-1

2

\newline

h(c)=1

for at least one

c

-1

2

Posted 3 months ago

Question

g

be a continuous function on the closed interval

[-3,3]

g(-3)=0

g(3)=6

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

g(c)=-2

for at least one

c

-3

3

\newline

g(c)=-2

for at least one

c

0

6

\newline

g(c)=5

for at least one

c

-3

3

\newline

g(c)=5

for at least one

c

0

6

Posted 3 months ago

Question

g

be a continuous function on the closed interval

[-1,4]

g(-1)=-4

g(4)=1

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

g(c)=3

for at least one

c

-1

4

\newline

g(c)=-3

for at least one

c

-4

1

\newline

g(c)=3

for at least one

c

-4

1

\newline

g(c)=-3

for at least one

c

-1

4

Posted 3 months ago

Question

g

be a continuous function on the closed interval

[-1,3]

g(-1)=-2

g(3)=-5

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

g(c)=-3

for at least one

c

-5

-2

\newline

g(c)=0

for at least one

c

-5

-2

\newline

g(c)=0

for at least one

c

-1

3

\newline

g(c)=-3

for at least one

c

-1

3

Posted 3 months ago

Question

f

be a continuous function on the closed interval

[-2,1]

f(-2)=3

f(1)=6

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

f(c)=0

for at least one

c

-2

1

\newline

f(c)=0

for at least one

c

3

6

\newline

f(c)=4

for at least one

c

-2

1

\newline

f(c)=4

for at least one

c

3

6

Posted 3 months ago

Question

f

be a continuous function on the closed interval

[-5,0]

f(-5)=0

f(0)=5

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

f(c)=-2

for at least one

c

-5

0

\newline

f(c)=2

for at least one

c

0

5

\newline

f(c)=2

for at least one

c

-5

0

\newline

f(c)=-2

for at least one

c

0

5

Posted 3 months ago

Question

f

be a continuous function on the closed interval

[1,5]

f(1)=1

f(5)=-3

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

f(c)=2

for at least one

c

1

5

\newline

f(c)=-2

for at least one

c

-3

1

\newline

f(c)=-2

for at least one

c

1

5

\newline

f(c)=2

for at least one

c

-3

1

Posted 3 months ago

Question

h(x)=\frac{1}{x}

\newline

Can we use the intermediate value theorem to say the equation

h(x)=0.5

has a solution where

-1 \leq x \leq 1

\newline

1

\newline

(A) No, since the function is not continuous on that interval.

\newline

0

5

h(-1)

h(1)

\newline

(C) Yes, both conditions for using the intermediate value theorem have been met.

Posted 3 months ago

Question

h

be a continuous function on the closed interval

[0,4]

h(0)=2

h(4)=-2

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

h(c)=3

for at least one

c

0

4

\newline

h(c)=-1

for at least one

c

0

4

\newline

h(c)=-1

for at least one

c

-2

2

\newline

h(c)=3

for at least one

c

-2

2

Posted 3 months ago