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Irina18 [472]
2 years ago
9

Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.

Mathematics
1 answer:
Veseljchak [2.6K]2 years ago
4 0

This equation is separable, as

\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx

Integrate both sides; on the left, expand the fraction as

\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)

Then

\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C

\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C

Since y(0)=1, we get

\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1

so that the particular solution is

\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}

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Step-by-step explanation:

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Find the original price of a pair of shoes if the sale price is 9$ after a 75% discount
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Original price=$12

Step-by-step explanation:

First set that up as an equation

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The question wants to solve the following base on the data in your problem and the answer are the following:
#1 The graph has the Y axis that is labeled as the value of p(t) and the x is the value of T. The graph that has a line that connect the A and B which is decreasing in the value of P(t) while increasing the value of T.
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5 0
3 years ago
3/2 x 4/3 x 5/4… x 2006/2005
Lady_Fox [76]

Answer:

1003

Step-by-step explanation:

The problem is a classic example of a telescoping series of products, a series in which each term is represented in a certain form such that the multiplication of most of the terms results in a massive cancelation of subsequent terms within the numerators and denominators of the series.

The simplest form of a telescoping producta_{k} \ = \ \displaystyle\frac{t_{k}}{t_{k+1}}, in which the products of <em>n</em> terms is

a_{1} \ \times \ a_{2} \ \times \ a_{3} \ \times \ \cdots \times \ a_{n-1} \ \times \ a_{n} \ = \ \displaystyle\frac{t_{1}}{t_{2}} \ \times \ \displaystyle\frac{t_{2}}{t_{3}} \ \times \ \displaystyle\frac{t_{3}}{t_{4}} \ \times \ \cdots \ \times \ \displaystyle\frac{t_{n-1}}{t_{n}} \ \times \ \displaystyle\frac{t_{n}}{t_{n+1}} \\ \\ \-\hspace{5.55cm} = \ \displaystyle\frac{t_{1}}{t_{n+1}}..

In this particular case, t_{1} \ = \ 2 , t_{2} \ = \ 3, t_{3} \ = \ 4, ..... , in which each term follows a recursive formula of t_{n+1} \ = \ t_{n} \ + \ 1. Therefore,

\displaystyle\frac{t_{2}}{t_{1}} \times \displaystyle\frac{t_{3}}{t_{2}} \times \displaystyle\frac{t_{4}}{t_{3}} \times \cdots \times \displaystyle\frac{t_{n}}{t_{n-1}} \times \displaystyle\frac{t_{n+1}}{t_{n}} \ = \ \displaystyle\frac{3}{2} \times \displaystyle\frac{4}{3} \times \displaystyle\frac{5}{4} \times \cdots \times \displaystyle\frac{2005}{2004} \times \displaystyle\frac{2006}{2005} \\ \\ \-\hspace{5.95cm} = \ \displaystyle\frac{2006}{2} \\ \\ \-\hspace{5.95cm} = 1003

6 0
2 years ago
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