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arsen [322]
3 years ago
5

Let y = f(x be the solution to the differential equation dy/dx = x y with the intial condition f(1=2. what is the approximation

for f(2 if euler's method is used, starting at x=1, with a step size of 0.5
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
6 0
Denote the ODE by

\dfrac{\mathrm dy}{\mathrm dx}=g(x,y)=xy

For this equation, you'll be using the recurrence relation

\begin{cases}y_{n+1}=y_n+hg(x_n,y_n)\\x_{n+1}=x_n+h\\x_0=1\\y_0=2\\h=0.5\end{cases}

You have

f(1.5)\approx y_1=y_0+0.5x_0y_0=2+0.5(1)(2)=3
f(2)\approx y_2=y_1+0.5x_1y_1=3+0.5(1.5)(3)=5.25
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Can you help me out on this question?
astra-53 [7]

Answer:

4. x= 18

Step-by-step explanation:

We solve the problem like this:

Since 3 is proportional to 6 and 9 is proportional to x, we set up a proportion.

3/9 = 6/x  

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10. Simplify the rational expression by rationalizing the denominator. 3 sqrt 160/sqrt 1350x A) 2 sqrt 15x/15x B) 3 sqrt 160/135
Viefleur [7K]
The simplified expression by rationalizing the denominator is (C)\frac{4 \sqrt{15x} }{15x}.

First we must simplify the expression:
\frac{3 \sqrt{160} }{\sqrt{1350x} } =  \frac{12 \sqrt{10} }{15 \sqrt{6x} } =  \frac{4 \sqrt{10} }{5 \sqrt{6x} }

Then we factor the rational parts and cancel it out:
\frac{4 \sqrt{2} \sqrt{5} }{5 \sqrt{2} \sqrt{3x} } = \frac{4\sqrt{5} }{5\sqrt{3x} }

Then we rationalize the expression:
\frac{4\sqrt{5} }{5\sqrt{3x} } * \frac{\sqrt{3x} }{\sqrt{3x} } = \frac{4 \sqrt{15x} }{5*3x} = \frac{4 \sqrt{15x} }{15x}

<span>Finally, the simplified expression by rationalizing the denominator is (C)\frac{4 \sqrt{15x} }{15x}.</span>
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Unit activity: exponential and logarithmic functions
nirvana33 [79]

We will conclude that:

  • The domain of the exponential function is equal to the range of the logarithmic function.
  • The domain of the logarithmic function is equal to the range of the exponential function.

<h3>Comparing the domains and ranges.</h3>

Let's study the two functions.

The exponential function is given by:

f(x) = A*e^x

You can input any value of x in that function, so the domain is the set of all real numbers. And the value of x can't change the sign of the function, so, for example, if A is positive, the range will be:

y > 0.

For the logarithmic function we have:

g(x) = A*ln(x).

As you may know, only positive values can be used as arguments for the logarithmic function, while we know that:

\lim_{x \to \infty} ln(x) = \infty \\\\ \lim_{x \to0} ln(x) = -\infty

So the range of the logarithmic function is the set of all real numbers.

<h3>So what we can conclude?</h3>
  • The domain of the exponential function is equal to the range of the logarithmic function.
  • The domain of the logarithmic function is equal to the range of the exponential function.

If you want to learn more about domains and ranges, you can read:

brainly.com/question/10197594

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