dy/dx-y=2ex
d(exy)/dx=2e2x
exy=e2x+c
f(0)=0..
e0(o)=e2*0+c
c=-1
therefore f(x)=e2x-1
ex
Now...
A differentiable function satisfies f'(x)=f(x)+2ex with initial conditions f(0) =0. The area enclosed by f(x) and the x axis is____
dy/dx-y=2ex
d(exy)/dx=2e2x
exy=e2x+c
f(0)=0..
e0(o)=e2*0+c
c=-1
therefore f(x)=e2x-1
ex
Now...
yes abhi i have also got the same eqn .tx for ur reply da
i asked only to confirm
area between f(x) and -ve x axix will be finite and equal to 1
y = 2xex
f'(x) = f(x) + 2ex
f'(x) - f(x) = 2ex
f'(x)e-x - f(x)e-x = 2
d/dx(f(x)e-x) = 2
rest is simple