Question:
Consider the differential equation, \(y\,dx-\left(x+y^{2}\right)dy=0\). If for \(y=1\), \(x\) takes value 1, then value of \(x\) when \(y=4\) is :
Consider the differential equation, \(y\,dx-\left(x+y^{2}\right)dy=0\). If for \(y=1\), \(x\) takes value 1, then value of \(x\) when \(y=4\) is :
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Treat \(x\) as a function of \(y\): the equation becomes linear in \(x\) with integrating factor \(1/y\).
Updated On: Nov 12, 2026
- \(9\)
- \(16\)
- \(36\)
- \(64\)
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The Correct Option is B
Solution and Explanation
Rewrite: \(\frac{dx}{dy}=\frac{x+y^{2}}{y}\), i.e. \(\frac{dx}{dy}-\frac{x}{y}=y\).
This is linear in \(x\) with integrating factor \(e^{-\int dy/y}=\frac{1}{y}\).
Hence \(\frac{d}{dy}\left(\frac{x}{y}\right)=1\), so \(\frac{x}{y}=y+c\).
Using \(x=1\) at \(y=1\): \(1=1+c\Rightarrow c=0\), so \(x=y^{2}\).
At \(y=4\): \(x=16\).
Answer: (2) 16
This is linear in \(x\) with integrating factor \(e^{-\int dy/y}=\frac{1}{y}\).
Hence \(\frac{d}{dy}\left(\frac{x}{y}\right)=1\), so \(\frac{x}{y}=y+c\).
Using \(x=1\) at \(y=1\): \(1=1+c\Rightarrow c=0\), so \(x=y^{2}\).
At \(y=4\): \(x=16\).
Answer: (2) 16
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