Question

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

Answer 1

The solution is
`y = - ln((5)/(2)x^(2) + C)`

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

`(dy)/(dx) + 5xe^(y) = 0`

Then,

`(dy)/(dx) = - 5xe^(y)`

`(dy)/(e^(y)) = - 5x dx`

Then, we integrate both sides

`\int {(dy)/(e^(y))} =\int {- 5x dx}`

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

Then,

`-e^(-y) = -(5)/(2)x^(2) + C`

`e^(-y) = (5)/(2)x^(2) + C`

Then,

`ln(e^(-y)) = ln((5)/(2)x^(2) + C)`

Then,

`-y = ln((5)/(2)x^(2) + C)`

Hence,

`y = - ln((5)/(2)x^(2) + C)`