Solve the initial value problem.
-$$\frac { d y } { d x } = 12 ( 4 x - 5 ) ^ { - 6 } , \quad y ( 0 ) = 1$$
A) $y = - \frac { 3 } { 5 } ( 4 x - 5 ) ^ { - 5 } + 1$
B) $y = ( 4 x - 5 ) ^ { - 5 } - 5$
C) $y = - \frac { 12 } { 5 } ( 4 x - 5 ) ^ { - 5 } + 1$
D) $y = - \frac { 3 } { 7 } ( 4 x - 5 ) ^ { - 7 } + 1$

Question

Quizplus · Accepted Answer

To solve the initial value problem, integrate the given derivative $\frac{dy}{dx} = 12(4x - 5)^{-6}$. The integral of $12(4x - 5)^{-6}$ with respect to $x$ is $-\frac{3}{5}(4x - 5)^{-5} + C$. Using the initial condition $y(0) = 1$, we find $C = 1$, leading to the solution $y = -\frac{3}{5}(4x - 5)^{-5} + 1$.

Solve the Initial Value Problem dydx=12(4x−5)−6,y(0)=1\frac { d y } { d x } = 12 ( 4 x - 5 ) ^ { - 6 } , \quad y ( 0 ) = 1dxdy​=12(4x−5)−6,y(0)=1

Solve the Initial Value Problem $\frac { d y } { d x } = 12 ( 4 x - 5 ) ^ { - 6 } , \quad y ( 0 ) = 1$