Find the solution of the initial-value problem $y ^ { \prime } = \frac { \ln x } { x y }$ , $y ( 1 ) = 2$ .
A) $y = \frac { 1 + x } { 1 + \ln x }$
B) $y = \frac { 8 x } { ( 1 + x ) ^ { 2 } }$
C) $y = 2 + 2 \ln x$
D) $y = \sqrt { 4 + ( \ln x ) ^ { 2 } }$
E) $y = x \ln x + 2 x$
F) $y = x \left( 1 + x ^ { 2 } \right)$
G) $y = x + \sqrt { 1 + \ln x }$
H) $y = \sqrt { x } ( 1 + x )$

Question

Find the solution of the initial-value problem $y ^ { \prime } = \frac { \ln x } { x y }$ , $y ( 1 ) = 2$ .
A) $y = \frac { 1 + x } { 1 + \ln x }$ 
B) $y = \frac { 8 x } { ( 1 + x ) ^ { 2 } }$ 
C) $y = 2 + 2 \ln x$ 
D) $y = \sqrt { 4 + ( \ln x ) ^ { 2 } }$ 
E) $y = x \ln x + 2 x$ 
F) $y = x \left( 1 + x ^ { 2 } \right)$ 
G) $y = x + \sqrt { 1 + \ln x }$ 
H) $y = \sqrt { x } ( 1 + x )$

Quizplus · Accepted Answer

To solve the initial-value problem $y' = \frac{\ln x}{xy}$, we separate variables and integrate both sides. The solution that satisfies the initial condition $y(1) = 2$ is $y = \sqrt{4 + (\ln x)^2}$. This is because integrating both sides leads to a function of $y$ that, when squared, includes a constant term (from the initial condition) and a variable term ($(\ln x)^2$), which matches the form of option D.

Find the Solution of the Initial-Value Problem y′=ln⁡xxyy ^ { \prime } = \frac { \ln x } { x y }y′=xylnx​

Find the Solution of the Initial-Value Problem $y ^ { \prime } = \frac { \ln x } { x y }$