2 years ago

Is this a function F(x)=(x+4)^2

Mathematics

1 answer:

Levart [38]2 years ago

7 0

No, it is not try factoring

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Answer:

The solution is $y = - ln(\frac{5}{2}x^{2} + C)$

Step-by-step explanation:

To solve the differential equation, we will find $y$

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That is, $y' + 5xe^{y} = 0$

This can be written as

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$\frac{dy}{dx} = - 5xe^{y}$

$\frac{dy}{e^{y}} = - 5x dx$

Then, we integrate both sides

$\int {\frac{dy}{e^{y}}} =\int {- 5x dx}$

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Then,

$-e^{-y} = -\frac{5}{2}x^{2} + C$

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$-y = ln(\frac{5}{2}x^{2} + C)$

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$y = - ln(\frac{5}{2}x^{2} + C)$

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