Question:
The general solution of \(e^{\frac{dy}{dx}}=x\) is
The general solution of \(e^{\frac{dy}{dx}}=x\) is
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Take the natural logarithm of both sides to get dy by dx, then integrate by parts.
Updated On: Nov 12, 2026
- \(y=x\log x-x+c\)
- \(y=x\log x+c\)
- \(y=1/x+c\)
- \(y=x.\,\log x+x+c\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Take the logarithm to isolate \(\frac{dy}{dx}\).
Step 2: Isolate the derivative
\(e^{dy/dx}=x\Rightarrow\frac{dy}{dx}=\log x\).
Step 3: Integrate
\(y=\int\log x\,dx\). By parts with \(u=\log x\) and \(dv=dx\): \(y=x\log x-\int x\cdot\frac{1}{x}dx\).
Step 4: Result
\(y=x\log x-x+c\).
Final Answer:
The general solution is y = x log x − x + c. \[ \boxed{\text{(A) }y=x\log x-x+c} \]
Take the logarithm to isolate \(\frac{dy}{dx}\).
Step 2: Isolate the derivative
\(e^{dy/dx}=x\Rightarrow\frac{dy}{dx}=\log x\).
Step 3: Integrate
\(y=\int\log x\,dx\). By parts with \(u=\log x\) and \(dv=dx\): \(y=x\log x-\int x\cdot\frac{1}{x}dx\).
Step 4: Result
\(y=x\log x-x+c\).
Final Answer:
The general solution is y = x log x − x + c. \[ \boxed{\text{(A) }y=x\log x-x+c} \]
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