Solve the initial value problem for y as a function of x.
-$$\sqrt { x ^ { 2 } - 4 } \frac { d y } { d x } = x , x 2 , y ( 4 ) = 0$$
A) $y = \sqrt { x ^ { 2 } - 4 } - 2 \sqrt { 3 }$
B) $y = \sqrt { x ^ { 2 } - 4 }$
C) $y = \frac { \sqrt { x ^ { 2 } - 4 } } { x }$
D) $y = \frac { \sqrt { x ^ { 2 } - 4 } } { x } - \frac { \sqrt { 3 } } { 2 }$

Question

Solve the initial value problem for y as a function of x.
-$$\sqrt { x ^ { 2 } - 4 } \frac { d y } { d x } = x , x > 2 , y ( 4 ) = 0$$
A) $y = \sqrt { x ^ { 2 } - 4 } - 2 \sqrt { 3 }$
B) $y = \sqrt { x ^ { 2 } - 4 }$
C) $y = \frac { \sqrt { x ^ { 2 } - 4 } } { x }$
D) $y = \frac { \sqrt { x ^ { 2 } - 4 } } { x } - \frac { \sqrt { 3 } } { 2 }$

Quizplus · Accepted Answer

To solve the initial value problem, we first separate variables and integrate both sides with respect to $x$. Given $\sqrt{x^2 - 4} \frac{dy}{dx} = x$, we can rewrite this as $\frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 4}}$. Integrating both sides with respect to $x$, we get $y = \int \frac{x}{\sqrt{x^2 - 4}} dx$. Using a substitution such as $u = x^2 - 4$, $du = 2x dx$, the integral simplifies to $\frac{1}{2} \int \frac{du}{\sqrt{u}}$, which evaluates to $\sqrt{u} + C = \sqrt{x^2 - 4} + C$. To find $C$, we use the initial condition $y(4) = 0$, which gives $0 = \sqrt{4^2 - 4} + C$, leading to $C = -2\sqrt{3}$. Therefore, the solution is $y = \sqrt{x^2 - 4} - 2\sqrt{3}$.

Solve the Initial Value Problem for Y as a Function x2−4dydx=x,x>2,y(4)=0\sqrt { x ^ { 2 } - 4 } \frac { d y } { d x } = x , x > 2 , y ( 4 ) = 0x2−4​dxdy​=x,x>2,y(4)=0

Solve the Initial Value Problem for Y as a Function $\sqrt { x ^ { 2 } - 4 } \frac { d y } { d x } = x , x > 2 , y ( 4 ) = 0$