Explore BrainMass

# Initial value - find the solution to initial-value problem

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Find the solution to initial-value problem:
(x^2+1)y'= xe^y with initial condition y(0)= -1.

https://brainmass.com/math/basic-calculus/initial-value-solution-initial-value-problem-192221

## SOLUTION This solution is FREE courtesy of BrainMass!

(x^2 + 1) y' = x e^y

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

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

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

Upon integration, we get

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

y(0) = -1

Therefore, - e^1 = (1/2) ln 1 + C

Thus, C = -e

Therefore, - e^-y = (1/2) ln (x^2 + 1) - e

e^-y = e - ln [sqrt(x^2 + 1)]

-y = ln [e - ln{sqrt(x^2 + 1)}]

y = - ln [e - ln{sqrt(x^2 + 1)}] is the solution.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!