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larisa [96]
4 years ago
15

Please help me pls 7tyh grade math

Mathematics
1 answer:
Andrews [41]4 years ago
8 0
(4/7)-(1/3)
(12/31)-(7/21)
(5/21) is left
Or
In decimal form
(4/7)=0.571
(1/3)= 0.333
0.571-0.333
=0.238
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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}}

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(it means your solution is right)

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