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# Licensed to the Apache Software Foundation (ASF) under one
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# distributed with this work for additional information
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# to you under the Apache License, Version 2.0 (the
# “License”); you may not use this file except in compliance
# with the License. You may obtain a copy of the License at
#
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# software distributed under the License is distributed on an
# “AS IS” BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
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# specific language governing permissions and limitations
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“””
Example AWS Python Client for EC2
“””
# Import Required Libraries
import boto3
from boto3.session import Session
# Instantiate an AWS Client
client = boto3.client(‘ec2′, region_name=’us-west-2’)
# Create a Session
session = Session(aws_access_key_id=”,
aws_secret_access_key=”,
aws_security_token=
)
# Get Credentials
credentials = session.get_credentials()
# Use the Client to List EC2 Instances
instances = client.describe_instances()
# List EC2 Instances
for instance in instances:
print(“ID: %s | Region: %s | AMI: %s” %
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You’re reading an article about Norton internet security 2017 key/ serial key/ license key and you can find articles Norton internet security 2017 key/ serial key/ license key this with the url If you want to read an article about Norton internet security 2017 key/ serial key/ license key, please click here.Q:
Find a time dependent solution to a scalar equation
I have $y’ = y^2$ where $y(0) = 10$, and I have to find a time dependent solution to this equation that is smaller than $10$, and also such that $t\in [0,20]$. How should I go about finding a solution to this?
The answer should be that $y = 10^{1/3}-10^{ -1/3}$.
A:
There is no solution. Note that
$$
y’ = \frac{1}{3} y^3
$$
and we get the solution
$$
y = \sqrt[3]{10} -\sqrt[3]{10} = 0.
$$
So the required solution is not defined in some time intervals.
The solution $y=10^{1/3}-10^{ -1/3}$ of your problem may be interpreted as follows. If we write the initial value problem as $y’ = f(y,t)$, with $y(0)=10$, then we have
$$
\frac{d}{dt} \left( f(y,t) – y^2 \right) = 0
$$
which means that
$$
\frac{d}{dt} \left( f(y,t) – y^2 \right) = 0 \quad \Longleftrightarrow \quad \frac{df}{dt} = 2 y f.
$$
In a sense, this formula is saying the following: If you have a solution $y(t)$ of the first-order equation