Answer:
The solution is [tex]y = - ln(\frac{5}{2}x^{2} + C)[/tex]
Step-by-step explanation:
To solve the differential equation, we will find [tex]y[/tex]
From the given equation, y' + 5xey = 0.
That is, [tex]y' + 5xe^{y} = 0[/tex]
This can be written as
[tex]\frac{dy}{dx} + 5xe^{y} = 0[/tex]
Then,
[tex]\frac{dy}{dx} = - 5xe^{y}[/tex]
[tex]\frac{dy}{e^{y}} = - 5x dx[/tex]
Then, we integrate both sides
[tex]\int {\frac{dy}{e^{y}}} =\int {- 5x dx}[/tex]
[tex]\int {e^{-y}dy }} =\int {- 5x dx}[/tex]
Then,
[tex]-e^{-y} = -\frac{5}{2}x^{2} + C[/tex]
[tex]e^{-y} = \frac{5}{2}x^{2} + C[/tex]
Then,
[tex]ln(e^{-y}) = ln(\frac{5}{2}x^{2} + C)[/tex]
Then,
[tex]-y = ln(\frac{5}{2}x^{2} + C)[/tex]
Hence,
[tex]y = - ln(\frac{5}{2}x^{2} + C)[/tex]
