Question

If the slope of the tangent to the curve at any point is equal to -y + e^-x, then the equation of the curve passing through the origin is ______  

Options

• y + xe^x = 0 

• y - xe^x = 0 

• ye^x + x = 0 

• ye^x - x = 0

Answer 1

If the slope of the tangent to the curve at any point is equal to -y + e^-x, then the equation of the curve passing through the origin is ye^x - x = 0.

Explanation:

`dy/dx = -y + e^-x`

∴ `dy/dx + y = e^-x`

∴ I.F. = `e^{int1.dx} = e^x`

∴ solution of the given equatiyon is

`ye^x = inte^x e^-x dx + c`

= x + c

∴ `ye^x - x = c` ..........(i)

Since the curve passes through origin (0, 0)

∴ 0 - 0 = c

⇒ c = 0

∴ `ye^x - x = 0` .........[From (i)]