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IceJOKER [234]
3 years ago
8

What is n-9 divided by 10=-2

Mathematics
1 answer:
DerKrebs [107]3 years ago
3 0

The value of n is 11.

<u>Step-by-step explanation</u>:

The given expression is (n-9) ÷ 10 = -2

<u>To find the value of n</u> :

  • From the expression, it is determined that (n-9) is the numerator and 10 is the denominator.
  • Cross multiply the denominator in the left side of the expression to the another side.

The expression becomes (n-9) = (10\times -2).

⇒ n-9 = -20

Keep the n term on one side and the constants on other side.

⇒ n = -20+9

⇒ n = -11

Therefore, the value of n is 11.

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Y''+y'+y=0, y(0)=1, y'(0)=0
mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

We get,

\left ( r^2+r+1 \right )e^{rt}=0

Since e^{rt}\neq 0, r^2+r+1=0

{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )

Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

Therefore,

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

3 0
3 years ago
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