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timurjin [86]
3 years ago
7

How do you know where to put the constant when finding general solutions for differential equations?

Mathematics
1 answer:
Norma-Jean [14]3 years ago
5 0
Let's suppose we want to solve y'=y with y(0)=2. Separating variables and integrating, we get

\displaystyle\int\dfrac{\mathrm dy}y=\int\mathrm dx\implies\ln|y|=x+C\implies y=e^{x+C}

Leaving the solution in this form, the initial condition gives

2=e^{0+C}=e^C\implies C=\ln2

This means the solution is y=e^{x+\ln2}.

Now if we were to write y=e^{x+C}=e^xe^C=Ce^x, then we would have found

2=Ce^0\implies C=2

so that the solution would have been y=2e^x.

But these two solutions are the same, since y=e^{x+\ln2}=e^xe^{\ln2}=2e^x. So we get the same solution regardless of where we place C, despite getting different values for C.
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krok68 [10]
The answer is 68 because is repeating 3 times
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Alex Ar [27]

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Step-by-step explanation:

straight line =180

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6 0
3 years ago
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Lieutenant James made an average of 3 arrests per week for 4 weeks. How many arrests does she need to make in the fifth week to
tankabanditka [31]

Answer:

5

Step-by-step explanation:

We want the average arrests to be 4. It can be done by calculating the average between the average arrests and arrests during the 5th week, so we get:

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Where Avg(x) is the average of arrests in given weeks, and A(x) is the number of arrests in a given week.

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3 0
3 years ago
The were 12 men at the block party and 40 woman. What fraction of the people were men
Rudiy27

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5 0
3 years ago
PLEASE ANSWER, WILL MARK CORECT ANSWER BRAINLIEST
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Answer:

B, D

Step-by-step explanation:

-3(0.15 - 0.2 +0.25p) = -3(-0.05 +0.25p) = 0.15-0.75p

A doesn't work because -3(0.15 - 0.2 +0.25p) = -0.45+0.6-0.75

C doesn't work because 3(0.15+0.2+0.25p) = 0.45+0.6+0.75 which doesn't equal -0.45+0.6-0.75.

E doesn't work because 0.15-0.2 = -0.05 not 0.05.

4 0
3 years ago
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