Question:
If \(y=y(x)\) is the solution of the differential equation
\( \frac{5+e^x}{2+y}\cdot\frac{dy}{dx}+e^x=0 \)
satisfying \(y(0)=1\), then a value of \(y(\log_e 13)\) is :
If \(y=y(x)\) is the solution of the differential equation
\( \frac{5+e^x}{2+y}\cdot\frac{dy}{dx}+e^x=0 \)
satisfying \(y(0)=1\), then a value of \(y(\log_e 13)\) is :
\( \frac{5+e^x}{2+y}\cdot\frac{dy}{dx}+e^x=0 \)
satisfying \(y(0)=1\), then a value of \(y(\log_e 13)\) is :
Show Hint
Separate the variables: \(\frac{dy}{2+y}=-\frac{e^x}{5+e^x}dx\) and integrate.
Updated On: Nov 5, 2026
- \(-1\)
- \(1\)
- \(0\)
- \(2\)
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The Correct Option is A
Solution and Explanation
Separate variables: \(\frac{dy}{2+y}=-\frac{e^x\,dx}{5+e^x}\).
Integrate: \(\ln|2+y|=-\ln(5+e^x)+\ln K\), i.e. \((2+y)(5+e^x)=K\).
Use \(y(0)=1\): \(K=(2+1)(5+1)=18\).
At \(x=\ln13\): \(e^x=13\), so \((2+y)(18)=18\Rightarrow 2+y=1\Rightarrow y=-1\).
Answer: (1) \(-1\)
Integrate: \(\ln|2+y|=-\ln(5+e^x)+\ln K\), i.e. \((2+y)(5+e^x)=K\).
Use \(y(0)=1\): \(K=(2+1)(5+1)=18\).
At \(x=\ln13\): \(e^x=13\), so \((2+y)(18)=18\Rightarrow 2+y=1\Rightarrow y=-1\).
Answer: (1) \(-1\)
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