Find the particular antiderivative that satisfies the following conditions: dy/dx=4x^(-2)+2x^(-1)-3; y(1)=6. y=?

Answer 1

Given: y' = 4x^(-2) + 2x^(-1) - 3

and y(1) = 6

Find: y

Take the integral of both sides.

Integral [ y' ] = Integral [ 4x^(-2) + 2x^(-1) - 3 ]

y = Integral [ 4x^(-2) + 2x^(-1) - 3 ]

y = (-4)x^(-1) + 2*ln|x| - 3x + C

y = -4/x + 2*ln|x| - 3x + C

Using (x,y) = (1,6)

6 = -4/1 + 2*ln|1| - 3(1) + C

6 = -4 + 0 - 3 + C

6 = -7 + C

13 = C

Therefore:

y = -4/x + 2*ln|x| - 3x + C

y = -4/x + 2*ln|x| - 3x + 13

Answer 2

y=∫4/x² +2/x -3dx = ∫4/x²dx +∫2/xdx - 3∫dx = -4/x +2ln(x) -3x + C

y(1) = -4/(1) + 2(0) -3 + C = 6 ; Note that ln(1) = 0

C = 7+6 = 13

y = -4/x +2ln(x) -3x + 13