How do you find the general solution to #dy/dx=xe^y#?

Question

Ezra Adams · Accepted Answer

#y = ln (2/ (C - x^2))# #y' = x e^y# #e^(-y)y' = x # #int \ e^(-y)y' \ dx =int \ x \ dx# #int \ e^(-y)\ dy =int \ x \ dx# #- e^(-y) =x^2/2 + C# #e^(-y) =C - x^2/2# #e^(y) = 1/ (C - x^2/2)# #= 2/ (C - x^2)# #y = ln (2/ (C - x^2))#

Luke Alvarez · Answer

undefined To find the general solution to the differential equation $\frac{dy}{dx} = x e^y$, we can separate variables and integrate both sides with respect to $x$ and $y$ respectively. Here's how:

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

Separate variables:

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

Integrate both sides:

$$
\int \frac{1}{e^y} \, dy = \int x \, dx
$$

The integral of $1/e^y$ can be evaluated as $-e^{-y}$, and the integral of $x$ with respect to $x$ is $\frac{1}{2}x^2$:

$$
-e^{-y} = \frac{1}{2}x^2 + C
$$

where $C$ is the constant of integration.

To find $y$, we can solve for $y$:

$$
e^{-y} = -\frac{1}{2}x^2 + C
$$

Taking the reciprocal of both sides:

$$
e^y = -\frac{1}{-\frac{1}{2}x^2 + C}
$$

$$
e^y = \frac{1}{\frac{1}{2}x^2 - C}
$$

$$
y = \ln\left(\frac{1}{\frac{1}{2}x^2 - C}\right)
$$

So, the general solution to the differential equation $\frac{dy}{dx} = x e^y$ is $y = \ln\left(\frac{1}{\frac{1}{2}x^2 - C}\right)$, where $C$ is an arbitrary constant.