Find the general solution for the differential equation. Do not attempt to find an explicit expression for the solution.
\[
y^{\prime}=\frac{e^{2 x}}{\ln y}
\]
A. $\ln y-y=\frac{e^{x}}{2}+c$
B. $\ln y-y=\frac{e^{x}}{2}$
C. $y \ln y-y=\frac{e^{2 x}}{2}$
D. $y \ln y-y=\frac{e^{2 x}}{2}+c$
Answer
Comparing the form of the solution obtained with the options given in the question, the general solution for the differential equation is \(\boxed{y \ln y - y = \frac{e^{2x}}{2} + C}\), which corresponds to option D.
Step 1 ：Given the differential equation \(y^{\prime}=\frac{e^{2 x}}{\ln y}\).
Step 2 ：This is a first order differential equation. It can be solved by separating variables and integrating.
Step 3 ：Rewrite the equation as \(y' \ln y = e^{2x}\).
Step 4 ：Integrate both sides of the equation. The integral of the left side is \(y \ln y - y\) and the integral of the right side is \(\frac{e^{2x}}{2}\).
Step 5 ：The general solution should be in the form of \(y \ln y - y = \frac{e^{2x}}{2} + C\), where \(C\) is the constant of integration.
Step 6 ：Comparing the form of the solution obtained with the options given in the question, the general solution for the differential equation is \(\boxed{y \ln y - y = \frac{e^{2x}}{2} + C}\), which corresponds to option D.
