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Leni [432]
3 years ago
8

Solve the differential equation. y' + 5xey = 0.

Mathematics
1 answer:
Gwar [14]3 years ago
8 0

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

From the given equation, y' + 5xey = 0.

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

This can be written as

\frac{dy}{dx} + 5xe^{y} = 0

Then,

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

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

Then, we integrate both sides

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

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

Then,

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

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

Then,

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

Then,

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

Hence,

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

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