Question:
Let y = y(x) be the solution curve of the differential equation,
\( \left( {{y^2} - x} \right){{dy} \over {dx}} = 1 \), satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
Let y = y(x) be the solution curve of the differential equation,
\( \left( {{y^2} - x} \right){{dy} \over {dx}} = 1 \), satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
\( \left( {{y^2} - x} \right){{dy} \over {dx}} = 1 \), satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
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Treat x as a function of y: the equation becomes linear, \(\dfrac{dx}{dy}+x=y^2\); solve with integrating factor \(e^y\).
Updated On: Oct 31, 2026
- 2 + e
- -e
- 2
- 2 - e
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The Correct Option is D
Solution and Explanation
Step 1: \((y^2-x)\dfrac{dy}{dx}=1\Rightarrow\dfrac{dx}{dy}=y^2-x\Rightarrow\dfrac{dx}{dy}+x=y^2\).
Step 2: Integrating factor \(e^{y}\): \(xe^{y}=\int y^2e^{y}dy=e^{y}(y^2-2y+2)+C\).
Step 3: So \(x=y^2-2y+2+Ce^{-y}\). Using \(y(0)=1\), i.e. \(x=0\) at \(y=1\): \(0=1-2+2+Ce^{-1}\Rightarrow C=-e\).
Step 4: The curve meets the x-axis where \(y=0\): \(x=0-0+2-e\cdot e^{0}=2-e\).
Answer: (D) \(2-e\)
Step 2: Integrating factor \(e^{y}\): \(xe^{y}=\int y^2e^{y}dy=e^{y}(y^2-2y+2)+C\).
Step 3: So \(x=y^2-2y+2+Ce^{-y}\). Using \(y(0)=1\), i.e. \(x=0\) at \(y=1\): \(0=1-2+2+Ce^{-1}\Rightarrow C=-e\).
Step 4: The curve meets the x-axis where \(y=0\): \(x=0-0+2-e\cdot e^{0}=2-e\).
Answer: (D) \(2-e\)
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