Ask question

Ask question

All categories

2 years ago

13

Please help. I will give brainliest immediately.

1 answer:

DerKrebs [107]2 years ago

7 0

Answer:

I'm pretty sure it is d, 5/22

You might be interested in

Find the general solution of<br><br> dz/dt=5ze^(sint)cost+2e^(sint)cost

gizmo_the_mogwai [7]

Answer:

The general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

Step-by-step explanation:

The given differential equation is

$\frac{dz}{dt}=5ze^{\sin t}\cos t+2e^{\sin t}\cos t$

Taking out common factors.

$\frac{dz}{dt}=(5z+2)e^{\sin t}\cos t$

Using variable separable method, we get

$\frac{dz}{5z+2}=e^{\sin t}\cos t dt$

Integrate both sides.

$\int\frac{dz}{5z+2}=\int e^{\sin t}\cos t dt$

$I_1=I_2$ .... (1)

First solve LHS,

$I_1=\int\frac{dz}{5z+2}$

Substitute $5z+2=u$

$5\frac{dz}{du}=1$

$dz=\frac{1}{5}du$

$I_1=\frac{1}{5}\int\frac{du}{u}$

$I_1=\frac{1}{5}ln|u|+C_1$

$I_1=\frac{1}{5}ln|5z+2|+C_1$

Now, solve RHS,

$I_2=\int e^{\sin t}\cos t dt$

Substitute $\sin t=v$ ,

$\cos t\frac{dt}{dv}=1$

$\cos tdt=dv$

$I_2=\int e^{v}dv$

$I_2=e^{v}+C_2$

$I_2=e^{\sin t}+C_2$

Subtitle the values of I₁ and I₂ in equation (1).

$\frac{1}{5}ln|5z+2|+C_1=e^{\sin t}+C_2$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C_2-C_1$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C$

Therefore the general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

4 0

3 years ago

Please help!!!!!! ASAP!!!

Kitty [74]

Answer:

B. po ang answer sa palagay ko lang po

6 0

2 years ago

What is the answer and how do you find it? Show work or no points.

NNADVOKAT [17]

Answer:

Answer is D

Step-by-step explanation:

I kinda understand it but it need to be neat so yh next time you have aa problem like this just write it neat for people to understand cleary what is your question.

6 0

2 years ago

What is 27 over 40 as a decimal

baherus [9]

27 divided by 40 is equal to 0.675.

5 0

2 years ago

How do you do this question?

katen-ka-za [31]

Answer:

0

Step-by-step explanation:

∫ sin²(x) cos(x) dx

If u = sin(x), then du = cos(x) dx.

∫ u² du

⅓ u³ + C

⅓ sin³(x) + C

Evaluate between x=0 and x=π.

⅓ sin³(π) − ⅓ sin³(0)

0

7 0

3 years ago

Read 2 more answers