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3 years ago

What's number 5 please I need to know

Mathematics

1 answer:

Natali [406]3 years ago

5 0

Unit Rate is showing 1 pairs and what you can do to get the answer is 12/5.79

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Solve the linear differential equation 2xy' + y = 2√x

77julia77 [94]

Answer:

Step-by-step explanation:

General form of the linear differential equation can be written as:

$\frac{dy}{dx}+P(x)y=Q(x)$

For this case, we can rewrite the equation as:

$\frac{dy}{dx}+\frac{1}{2x}y=\frac{\sqrt{x}}{x}$

Here $P(x) =\frac{1}{2x}; Q(x)=\frac{\sqrt{x}}{x}$

To find the solution (y(x)), we can use the integration factor method:

$Fy(x)=\int Q(x)Fdx+C \rightarrow F=e^{\int P(x)dx$

Then $F=e^{\int \frac{1}{2x}dx}=e^{\frac{1}{2}\ln|x|\right}=\sqrt{|x|}$

So, we can find:

$y\sqrt{|x|}=\int \frac{\sqrt{x}\sqrt{|x|}}{x}dx+C$

Suppose that $x\in \double R$ , then $\sqrt{|x|}=\sqrt{x}$ , and we find:

$y\sqrt{x}=x+C \rightarrow y(x)=\sqrt{x}+\frac{C\sqrt{x}}{x}$

To check our solution is right or not, put your y(x) back to the ODE:

$y' = \frac{1}{2\sqrt{x}}-\frac{C}{2\sqrt{x^{3}}}$

$2xy'=\frac{x-C}{\sqrt{x}}$

$2xy'+y=\frac{x-C}{\sqrt{x}}+\sqrt{x}+\frac{C\sqrt{x}}{x}=2\sqrt{x}$

(it means your solution is right)

3 0

2 years ago

Peanuts are sold $1.67 for one pound. A bag of peanuts weights 2.5 pounds. How much will you pay for that one bag?

Gekata [30.6K]

Sym 10 times u bumbaclart idiaat

7 0

3 years ago

How could you find an equation of a quadratic function with the zeroes of -3 and 1?

Snowcat [4.5K]

Hello,

The simpliest quadratic function is y=(x+3)(x-1)
or y=x²+2x-3

6 0

3 years ago

How you would find the total area of the following composite shapes.

sergiy2304 [10]

Answer:

By breaking it up in shapes

Step-by-step explanation:

I annotated your photo and it's attached below :) that's an example on how to break shapes into easier parts. So yeah

Download pdf

7 0

3 years ago

Read 2 more answers

If f(x) = 2x-4 and g(x) = x2 - 4x, find g(-4) And the (x2 - 4x) the 2 is a exponent

Luden [163]

Evaluate the given function in x=-4.

This is, replace the x in the function for -4 and find the value of g(-4).

$\begin{gathered} g(x)=x^2-4x \\ g(-4)=(-4)^2-4(-4) \\ g(-4)=16+16 \\ g(-4)=32 \end{gathered}$

The value of g(-4) is 32.

5 0

1 year ago