I think it's the last one because as x increases by one, y increases by a constant rate of 5. The others don't increase by a constant rate. (I don't know if this is right but I think it is!)

5 0

3 years ago

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Asi a las 7:00 am.Una Persona cuenta un chisme a cuatro

Advocard [28]

Answer:4096

<h2>Step-by-step explanation4^6 </h2>

6 0

3 years ago

Solve the initial-value problem<br><br> y' = x^4 - \frac{1}{x}y, y(1) = 1.

natta225 [31]

The ODE is linear:

$y'=x^4-\dfrac yx$

$y'+\dfrac yx=x^4$

Multiplying both sides by $x$ gives

$xy'+y=x^5$

Notice that the left side can be condensed as the derivative of a product:

$(xy)'=x^5$

Integrating both sides with respect to $x$ yields

$xy=\dfrac{x^6}6+C$

$\implies y(x)=\dfrac{x^5}6+\dfrac Cx$

Since $y(1)=1$ ,

$1=\dfrac16+C\implies C=\dfrac56$

so that

$\boxed{y(x)=\dfrac{x^5}6+\dfrac5{6x}}$

4 0

3 years ago