Question

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

Answer 1

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,

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