JEE Mains · Maths · STD 12 - 9. differential equations
Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be the solution curve of the differential equation, \(\quad\left(y^{2}-x\right) \frac{d y}{d x}=1\) satisfying \(\mathrm{y}(0)=1 .\) This curve intersects the \(\mathrm{x}\) -axis at a point whose abscissa is
- A \(2+e\)
- B \(2\)
- C \(2-e\)
- D \(-e\)
Answer & Solution
Correct Answer
(C) \(2-e\)
Step-by-step Solution
Detailed explanation
\(\left(y^{2}-x\right) \frac{d y}{d x}=1\) \(\Rightarrow \frac{d x}{d y}+x=y^{2}\) I.F. \(=\mathrm{e}^{\int \mathrm{dy}}=\mathrm{e}^{\mathrm{y}}\) Solution is given by \(\mathrm{x} \mathrm{e}^{\mathrm{y}}=\int \mathrm{y}^{2} \mathrm{e}^{\mathrm{y}} \mathrm{dy}+\mathrm{C}\)…
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