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shtirl [24]
3 years ago
10

Dy/dx = y/x^2

Mathematics
1 answer:
Lady_Fox [76]3 years ago
8 0
Dy/dx = y/(x^2)
dy/y = dx/(x^2)
int[dy/y] = int[dx/(x^2)] ... apply integral to both sides
ln(|y|) = (-1/x) + C
|y| = e^{(-1/x) + C}
|y| = e^C*e^(-1/x)
|y| = C*e^(-1/x)
y = C*e^(-1/x)

So you have the correct answer. Nice job.

------------------------------------------------

Check:
y = C*e^(-1/x)
dy/dx = d/dx[C*e^(-1/x)]
dy/dx = d/dx[-1/x]*C*e^(-1/x)
dy/dx = (1/(x^2))*C*e^(-1/x)
is the expression for the left hand side (LHS)

y/(x^2) = [C*e^(-1/x)]/(x^2)
y/(x^2) = (1/(x^2))*C*e^(-1/x)
is the expression for the right hand side (RHS)

Since LHS = RHS, this confirms the solution for dy/dx = y/(x^2)
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When we compare the significance level \alpha=0.1 we see that p_v so we can reject the null hypothesis at 10% of significance. So the  the true mean is difference from 21 at this significance level.

Step-by-step explanation:

Data given and notation  

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p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the average age of the evening students is significantly different from 21, the system of hypothesis would be:  

Null hypothesis:\mu = 21  

Alternative hypothesis:\mu \neq 21  

The statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{23-21}{\frac{3.5}{\sqrt{49}}}=4    

P-value

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Conclusion  

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