The solution of the differential equation, e x ( x +1) dx +( ye y - xe x ) dy =0 with initial condition f(0)=0, is

Question

The solution of the differential equation, e x ( x +1) dx +( ye y - xe x ) dy =0 with initial condition f(0)=0, is (A)  x ex+2 y2 ey=0  (B) 2 x ex+y2 ey=0 (C) x ex-2 y2 ey=0 (D) 2 x ex-y2 ey=0

Tardigrade Admin · Accepted Answer

text put x ex=t (ex+x ex) (d x/d y)=(d t/d y) ∴ ( dt / dy )+( ye y - t )=0 ⇒ ( dt / dy )- t + ye y =0 I.F. e -∫ d y= e - y t ⋅ e - y =-∫ ye y e - y dy x e x e - y =-( y 2/2)+ C f(0)=0 ⇒ C =0 ; 2 xe x e - y + y 2=0

Q. The solution of the differential equation, $e^{x} (x + 1) d x + (y e^{y} - x e^{x}) d y = 0$ with initial condition $f (0) = 0$ , is

$x e^{x} + 2 y^{2} e^{y} = 0$

$2 x e^{x} + y^{2} e^{y} = 0$

$x e^{x} - 2 y^{2} e^{y} = 0$

$2 x e^{x} - y^{2} e^{y} = 0$

Q. The solution of the differential equation, ex(x+1)dx+(yey−xex)dy=0 with initial condition f(0)=0, is

xex+2y2ey=0

2xex+y2ey=0

xex−2y2ey=0

2xex−y2ey=0

put xex=t (ex+xex)dydx​=dydt​ ∴dydt​+(yey−t)=0⇒dydt​−t+yey=0 I.F. e−∫dy=e−y t⋅e−y=−∫yeye−ydy xexe−y=−2y2​+C f(0)=0⇒C=0;2xexe−y+y2=0

Q. The solution of the differential equation, $e^{x} (x + 1) d x + (y e^{y} - x e^{x}) d y = 0$ with initial condition $f (0) = 0$ , is

$x e^{x} + 2 y^{2} e^{y} = 0$

$2 x e^{x} + y^{2} e^{y} = 0$

$x e^{x} - 2 y^{2} e^{y} = 0$

$2 x e^{x} - y^{2} e^{y} = 0$