Differential Equation Entry?

[DrD]

How can this be solved using the prime: (Ideally, returning BOTH a general and a particular solution.)

(y*e^x) dx + (y^2 - 1) dy = 0, AND y(0) = 1

Desired results:
y^2 - ln(y^2) = -2*e^x + c1; {General Solution}
y^2 - ln(y^2) = -2*e^x + 3; {Particular solution}

[parisse]

desolve(y*exp(x)+(y^2-1)*y'=0) returns the implicit equation for the solutions (in display and if you interpret fsolve).
odesolve will solve this equation numerically, but not for this initial condition, because Cauchy-Lipschitz is not verified if y=1 (you can not express y' in terms of x and y).

[DrD]

(04-10-2018 11:53 AM)parisse Wrote:  desolve(y*exp(x)+(y^2-1)*y'=0) returns the implicit equation for the solutions (in display and if you interpret fsolve).
odesolve will solve this equation numerically, but not for this initial condition, because Cauchy-Lipschitz is not verified if y=1 (you can not express y' in terms of x and y).

Thanks, Bernard! I tried lots of other ways, but not that one, and I spent way too much time trying them!

-Dale-