Solve the initial value problem.
-$$\frac { d y } { d x } = \frac { 1 } { x + 5 } , y ( - 4 ) = 6$$
A) $y = \ln ( x + 5 ) + 6$
B) $y = - \frac { 1 } { ( x + 5 ) ^ { 2 } } + 6$
C) $y = - \frac { 1 } { ( x + 5 ) ^ { 2 } } + 7$
D) $y = \ln ( x + 5 ) + 6 - \ln 2$

Question

Quizplus · Accepted Answer

Integrating $\frac{dy}{dx} = \frac{1}{x+5}$ gives $y = \ln|x+5| + C$. Using $y(-4) = 6$, we find $C = 6$, so $y = \ln|x+5| + 6$. Since $x+5 > 0$ for the given domain, $y = \ln(x+5) + 6$.

Solve the Initial Value Problem dydx=1x+5,y(−4)=6\frac { d y } { d x } = \frac { 1 } { x + 5 } , y ( - 4 ) = 6dxdy​=x+51​,y(−4)=6

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