Given dy/dx = (1+y^2)e^x

To find the solution of the differential equation:

We write dy/dx = (1+y^2) e^x in the variable sparable form as below and integrate so that we get the primitive in the form y = f(x):

dy/(1+y^2) = e^x dx

Now we integrate both sides:

Int {dy/(1+y^2)} = Int e^x dx

arc tan y = e^x +c.

Arc tangent is an inverse function . So we can get y explicitly from this.

y = tan (e^x + c).

We have to solve dy/dx = (1+y^2)*e^x

First we will group all the terms containing y with dy on one side and those containing x with dx on the other.

dy/dx = (1+y^2)*e^x

=> dy / (1+ y^2) = e^x dx

=> dy / (1+ y^2) - e^x dx =0

now integrate both the side, we get

arc tan y – e^x = C

or in terms of y

=> arc tan y = e^x + C

=> y = tan (e^x + C)

**Therefore the result is y = tan (e^x + C)**