Question:
The solution of \(\log \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2 x-5 y, y(0)=0\) is
The solution of \(\log \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2 x-5 y, y(0)=0\) is
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Remove the logarithm, separate the variables and use the initial condition.
Updated On: Nov 13, 2026
\(2\mathrm{e}^{2x}+5\mathrm{e}^{5y}=6\)
\(5\mathrm{e}^{2x}-2\mathrm{e}^{5y}=3\)
\(2\mathrm{e}^{2x}-5\mathrm{e}^{5y}=6\)
\(5\mathrm{e}^{2x}+2\mathrm{e}^{5y}=3\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
\(\log\dfrac{dy}{dx}=2x-5y\) gives \(\dfrac{dy}{dx}=e^{2x}e^{-5y}\).
Step 2: Separate variables
\(e^{5y}dy=e^{2x}dx\).
Step 3: Integrate
\(\dfrac{e^{5y}}{5}=\dfrac{e^{2x}}{2}+C\).
Step 4: Use the condition
At \(x=0, y=0\): \(\dfrac{1}{5}=\dfrac{1}{2}+C\), so \(C=-\dfrac{3}{10}\). Multiplying by 10: \(2e^{5y}=5e^{2x}-3\), i.e. \(5e^{2x}-2e^{5y}=3\).
Final Answer:
The solution is \(5e^{2x}-2e^{5y}=3\). \[ \boxed{\text{(B) }5\mathrm{e}^{2x}-2\mathrm{e}^{5y}=3} \]
\(\log\dfrac{dy}{dx}=2x-5y\) gives \(\dfrac{dy}{dx}=e^{2x}e^{-5y}\).
Step 2: Separate variables
\(e^{5y}dy=e^{2x}dx\).
Step 3: Integrate
\(\dfrac{e^{5y}}{5}=\dfrac{e^{2x}}{2}+C\).
Step 4: Use the condition
At \(x=0, y=0\): \(\dfrac{1}{5}=\dfrac{1}{2}+C\), so \(C=-\dfrac{3}{10}\). Multiplying by 10: \(2e^{5y}=5e^{2x}-3\), i.e. \(5e^{2x}-2e^{5y}=3\).
Final Answer:
The solution is \(5e^{2x}-2e^{5y}=3\). \[ \boxed{\text{(B) }5\mathrm{e}^{2x}-2\mathrm{e}^{5y}=3} \]
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