Solve the initial value problem.
-$$\frac { d y } { d x } = \frac { 1 } { ( 2 + x ) ^ { 2 } } , y ( 0 ) = 2$$
A) $y = \frac { 1 } { 2 + x } + \frac { 3 } { 2 }$
B) $y = \frac { 1 } { 2 + x } + 2$
C) $y = - \frac { 1 } { 2 + x } + \frac { 5 } { 2 }$
D) $y = - \frac { 1 } { 2 + x } + 2$

Question

Quizplus · Accepted Answer

To solve the initial value problem, integrate $\frac{dy}{dx} = \frac{1}{(2+x)^2}$ to find $y$. The integral of $\frac{1}{(2+x)^2}$ with respect to $x$ is $-\frac{1}{2+x} + C$, where $C$ is the constant of integration. To find $C$, use the initial condition $y(0) = 2$, which gives $2 = -\frac{1}{2+0} + C$, leading to $C = \frac{5}{2}$. Therefore, the solution is $y = -\frac{1}{2+x} + \frac{5}{2}$.

Solve the Initial Value Problem dydx=1(2+x)2,y(0)=2\frac { d y } { d x } = \frac { 1 } { ( 2 + x ) ^ { 2 } } , y ( 0 ) = 2dxdy​=(2+x)21​,y(0)=2

Solve the Initial Value Problem $\frac { d y } { d x } = \frac { 1 } { ( 2 + x ) ^ { 2 } } , y ( 0 ) = 2$