The order and degree of the differential equation of all tangent lines to the parabola `y=x^2` is

Options

A. `y'''+3y''+3y'+y=0`
B. `y'''+3y''-3y'-y=0`
C. `y'''-3y''-3y'+y=0`
D. `y'''-3y''+3y'-y=0`

Answer

d

Explanation

Given ` y = ae^(x) + bx e^(x) + cx^(2)e^(x) " "` … (i)
On differentiating w.r.t x, we get
` y ' = ae^(x) + b (xe^(x)+e^(x)) + c (x^(2)e^(2)+2xe^(x))`
` rArr y ' (ae^(x) +bxe^(x) + cx^(2)e^(x)) + be^(x)+2cxe^(x)`
` rArr y' = y + be^(x) + 2 cxe^(x)" "` ...(ii)
Again , on differentiating w.r.t x , web get
` y'' = y' be^(x) + 2c (xe^(x)+e^(x))`
` rArr y'' = y' +be^(x) + 2cxe^(x) 2ce^(x)`
` rArr y '' = 2y' - y +2ce^(x)" "` [ from Eq.(ii) ] ...(iii)
And again . on differetniating w.r.t.x, we get
` y ''' = 2y'' - y' +2ce^(x)`
` rArr y''' = 2y'' - y' +(y'' - 2y'+y)` [ from Eq. (iii)]
` rArr y ''' - 3y'' + 3y' - y =0`