Explore BrainMass

Explore BrainMass

    Differential equations

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

    Solve

    (1-x^2)^(1/2)y'+1+y^2=0

    xy(1+x^2)y'-(1+y^2)=0

    xyy'=1+x^2+y^2+x^2y^2

    sinx(e^y + 1)dx=e^y(1+cosx)dy, Y(0)=0

    © BrainMass Inc. brainmass.com March 4, 2021, 5:36 pm ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/differential-equations-solved-2300

    Solution Preview

    1. (1-x^2)^(1/2)y'+1+y^2=0 implies that

    y'/(1+y^2)=-1/((1-x^2)^(1/2) ) (1)

    We have the formula

    The definite integral of

    1/((a^2+u^2) is 1/a* [ tan^(-1)(u/a)] +C

    The definite integral of

    1/((a^2-u^2)^(1/2) is sin^(-1)(u/a)+C.

    Integrate both sides of (1) to get ( with a=1)

    1/[ tan^(-1)(y)] = - sin^(-1)(x)+C.

    Thus

    tan(y)= 1/[- ...

    Solution Summary

    This showsh ow to solve several different differential equations.

    $2.19

    ADVERTISEMENT