Question

Find the equation of the curve that passes through (-2,-46) if its slope is given by

dy
dx
12x²2 - 8x for each x. Thank

Answer 1

Given the differential equation

`(dy)/(dx) = 12x^2 - 8x`

integrating both sides with respect to
`x` yields

`\displaystyle \int (dy)/(dx) \, dx = \int (12x^2 - 8x) \, dx`

`y = 4x^3 - 4x^2 + C`

Use the given point to solve for the constant
`C` :

`-46 = 4(-2)^3 - 4(-2)^2 + C \implies C = 2`

Then the equation of the curve is

`\boxed{y = 4x^3 - 4x^2 + 2}`

On the off-chance you instead meant something like

`(dy)/(dx) = (12x^2)/(2 - 8x) = -\frac{3x}2 - \frac38 + \frac3{8(1-4x)}`

integrating would instead give

`y = -\frac{3x^2}4 - \frac{3x}8 - \frac3{32} \ln|1-4x| + C`

Then

`-46 = -\frac94 - \frac3{32} \ln(9) + C \implies C = \frac3{32}\ln(9) - \frac{175}4`

and the particular solution follows. (But I suspect this is *not* what you meant.)