Solving A Nice Functional Equation

[@manojsurya1005]

Wow ,thats a new approach for me Great work

[@SyberMath]

Glad to hear that! ❤️

[@trnfncb11]

Set y=0 to get f(x)=f(x)f(0), i.e. f(0)=1. Now set y=-x, to get f(0)=1=e^(-2x^2)f(x)f(-x). So f(x)f(-x)=e^(2x^2). A solution to this is clearly f(x)=e^(x^2+kx) for any k.

[@RashmiRay-c1y]

ln f(x+y) = 2xy + lnf(x) + lnf(y).. Differentiating w.r.t. x, 1/f(x+y) partial_xf(x+y) = 2y + 1/f(x) df(x)/dx. Differentiating w.r.t. y, 1/f(x+y) partial_yf(x+y) = 2x + 1/f(y) df(y)/dy. But partial_xf(x+y) = partial_yf(x+y). Therefore, we get 2x + 1/f(y) df(y)/dy = 2y + 1/f(x) df(x)/dx or 1/f(x) df(x)/dx -2x = 1/f(y) df(y)/dy - 2y = k, a constant as the LHS is a function of x and the RHS is a function of y. Thus, 1/f(x) df(x)/dx -2x = k > d/dx(ln f(x) - x^2) = k > ln f(x) - x^2 = kx + l, where l is a constant. So, lnf(x) = x^2 + kx + l > f(x) = C e^(x^2 + kx), where C = e^l = a constant.

[@barakathaider6333]

👍

[@Nikos_Iosifidis]

Very interesting equation

[@SyberMath]

Glad you think so!

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@davidrojas5087]

Nice

[@SyberMath]

Thanks

[@wesleydeng71]

Let f(x) = g(x)*e^(x^2) and solution will follow quickly.

[@SyberMath]

Good thinking!

[@Anmol_Sinha]

Darn. This is so goood

[@robot8324]

Nice

[@SyberMath]

Thanks