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2 years ago

7

Tính đạo hàm y'(1)của hàm ẩn y=y(x) được cho bởi phương trình

1 answer:

galina1969 [7]2 years ago

8 0

Answer:

..gufwfjudgjbsrjjvxe djhse h hsehhf rhf4(&&-++(34

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Whats the capacity of a water bottle thats 8 inches

suter [353]

1 pt; or 1 pint seems most reasonable here.

8 0

3 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

3 years ago

The ratio of girls to boys in a a gardening club was 6:2 there were 16 boy how many are girls

marishachu [46]

Answer:

I don't know

Step-by-step explanation:

tdhdujdjbd

3 0

1 year ago

Need help with a math question

krek1111 [17]

Answer:

x=13°

Step-by-step explanation:

If BE is an angle bisector, then it divides the angle into two equal angles. This means that

∠ABE=∠EBC

Since ∠ABE=2x+20 and ∠EBC=4x-6, we have

2x+20=4x-6

2x-4x=-6-20

-2x=-26

2x=26

x=13°

4 0

3 years ago

Please answer quick!!

rewona [7]

Answer: B

Step-by-step explanation:

$\frac{2}{7k}(k-7)$

$\frac{2k-14}{7k}$

$\frac{2k}{7k}-\frac{14}{7k}$

$\frac{2}{7}-\frac{2}{k}$

7 0

3 years ago