Question

Find the tangents to the curve y=( x3 – 1)(x – 2) at the points where the curve cuts the x-axis

Answer 1

hope it will help you.

Answer 2

Step-by-step explanation:

The differential equation x (x-1)dy/dx-(x-2)y = x³(2x-1) can be solved in the following ways:

As per the question,

x (x-1)dy/dx - (x-2)y = x³(2x-1)

Dividing both sides by x(x-1), we get,

dy / dx - (x-2) / x(x-1) = x²(2x-1) / (x-1)

Let, a = e^{\int\limits {P} \, dx }e

∫Pdx

, where P = -(x-2) / x(x-1)

Now, Solving {\int\limits {P} \, dx }∫Pdx :

= - ∫ [(x-1) + (-1)] / x(x-11)

= ∫ - 1/ x + ∫ 1 / (x² - x)

= - ln x + ln (x-1) / x

= ln (x-1)/x²

Therefore we get = e^{ ln (x-1) / x^{2} }e

ln(x−1)/x

2

= (x-1)/x²

Now, y * a = x²(2x-1) / (x-1) * a

y * (x-1)/x² = ∫ x²(2x-1) / (x-1) * (x-1)/x²

= x² - x + C

Therefore, y = x² / (x-1) * ( x² - x + C )