# Solve the differential equation dy/ dx = (1 + y^2) * e^x .

*print*Print*list*Cite

### 2 Answers

### User Comments

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

Now we can write this in a form where all terms with y are grouped with dy and all terms with x are grouped with dx.

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

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

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

We'll integrate the first part with respect to y and the second part with respect to x.

=> Int [dy / (1 + y^2)] - Int [e^x dx] = C

We know that the integral of 1/ (1 + y^2) is arc tan y and the integral of e^x is e^x

=> arc tan y - e^x = C

=> arc tan y = C + e^x

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

**Therefore the required solution is y = tan (e^x + C).**

To solve the given expression, we'll re-write it:

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

Now, we'll have to integrate both sides, to calculate y:

Int dy = Int [(1 + y^2) * e^x]*dx

We'll integrate with respect to x, to the right side of the equation, so the sum 1 + y^2 will be considered as a constant:

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

We'll divide by (1 + y^2):

y/(1 + y^2) = e^x + C

y = e^x + y^2*e^x + C + Cy^2

y^2(C + e^x) - y + e^x + C= 0

The equation has solutions if it' discriminant is positive or zero:

delta = b^2 - 4ac

a = C + e^x

b = -1

c = e^x + C

delta = 1 - 4(e^x + C)^2

delta>0

1 - 4(e^x + C)^2>0

4(e^x + C)^2 < 1

(e^x + C)^2 < 1/4

e^x + C < +/- 1/2

e^x < C + 1/2

ln e^x < ln(C+/- 1/2)

**x < ln (C+/- 1/2)**