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3 years ago

Solve for z, express answer in standard form: 3iz + (2-4i) = (1+2i)z - 3i

Mathematics

1 answer:

zaharov [31]3 years ago

5 0

Hope this helps you!

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y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

The cost of renting a chain saw is $3.90 per hour plus $6.50 for a cán

puteri [66]

35.75
explanation-
3.90 ph x 7.5 hours = 29. 25
plus the gas can 6.50
29.25+6.50= 35.75

5 0

3 years ago

Which scatterplot would have a trend line with a negative slope

I am Lyosha [343]

Answer:

The middle one

Step-by-step explanation:

8 0

3 years ago

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Carrie and brent leave home at the same time. carrie drives due east at a rate of 30 mph, while brent drives due south at a rate

Helga [31]

<h3>They are 268 miles far apart after 4 hours</h3>

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem !

<u>Given :</u>

v₁ = 30 mph due east

v₂ = 60 mph due south

t = 4 hours

<u>Unknown :</u>

displacement = d = ?

<u>Solution :</u>

$d_1 = v_1 \times t$

$d_1 = 30 ~ mph \times 4 ~ h$

$d_1 = \boxed {120 ~ miles}$

$d_2 = v_2 \times t$

$d_2 = 60 ~ mph \times 4 ~ h$

$d_2 = \boxed {240 ~ miles}$

$d^2 = (d_1)^2 + (d_2)^2$

$d^2 = 120^2 + 240^2$

$d^2 = 72000$

$d = \sqrt{72000}$

$d = 120\sqrt{5} ~ miles$

$d \approx \boxed {268 ~ miles}$

<h2>Conclusion :</h2><h3>They are 268 miles far apart after 4 hours</h3>

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

6 0

3 years ago

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3. Determine whether the set of numbers can be the measures of the sides of a triangle. If so, classify the triangle as acute, o

julia-pushkina [17]

Answer:

all of them are acute

.......

3 0

3 years ago

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