Solve this: Q 49 Let f : R→R, such that f'(0) = 1 and f (x + y) = - Maths -

Question

Solve this:
Q 49 Let f   :   R → R , such that f'(0) = 1 and f
(x + y) = f (x) + f (y) + e x + y x + y - x e x - y e y + 2 x y
  ∀   x ,   y ∈ R then determine f (x)

Aarushi Mishra · Accepted Answer

fx+y=fx+fy+ex+yx+y-xex-yey+2xyPartially differentiate with respect to y∂ ∂yfx+y=∂ ∂yfx+∂ ∂yfy+∂ ∂yex+yx+y-∂ ∂yxex-∂ ∂yyey+2∂ ∂yxyf'x+y=0+f'y+ex+y∂ ∂yx+y+x+y∂ ∂yex+y-0-ey∂ ∂yy-y∂ ∂yey+2x∂ ∂yyf'x+y=f'y+ex+y+x+yex+y-ey-yey+2xPut y=0f'x=f'0+ex+xex-1+2xGiven f'0=1f'x=1+ex+xex-1+2xf'x=ex+xex+2xfx=∫f'xdx=∫ex+xex+2xdx=∫x+1exdx+2∫xdxNote dxdx=1Using∫fx+f'xexdx=exfx+cfx=xex+x2+cWe hadfx+y=fx+fy+ex+yx+y-xex-yey+2xyPut x=y=0f0=f0+f0+0-0-0+0f0=2f0f0=0Put x=0 in fx=xex+x2+cf0=0+0+c=0⇒c=0fx=xex+x2

Solve this: Q.49. Let f : R → R , such that f'(0) = 1 and f (x + y) = f (x) + f (y) + e x + y x + y - x e x - y e y + 2 x y ∀ x , y ∈ R then determine f (x).

Solve this:
Q.49. Let $f : R \to R$ , such that f'(0) = 1 and f (x + y) = f (x) + f (y) + $e^{x + y} (x + y) - x e^{x} - y e^{y} + 2 x y \forall x, y \in R$ then determine f (x).