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JEE Mains · Maths · STD 12 - 9. differential equations

ધારો કે વિકલ સમીકરણ \(\left(4+x^{2}\right) d y-2 x\left(x^{2}+3 y+4\right) d x=0\) નો ઉકેલ વક્ર \(y=y(x)\) એ ઉગમબિંદુ માંથી પસાર થાય છે.તો \(y(2)=\dots\dots\dots\)

  1. A \(10\)
  2. B \(11\)
  3. C \(12\)
  4. D \(13\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(12\)

Step-by-step Solution

Detailed explanation

\(\left(4+x^{2}\right) d y-2 x\left(x^{2}+3 y+4\right) d x\) \(\left(x^{2}+4\right) \frac{d y}{d x}=2 x^{3}+6 x y+8 x\) \(\left(x^{2}+4\right) \frac{d y}{d x}-6 x y=2 x^{3}+8 x\) \(\frac{d y}{d x}-\frac{6 x}{x^{2}+4} y=\frac{2 x^{3}+8 x}{x^{2}+y}\) L.I.…
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