JEE Mains · Maths · STD 12 - 9. differential equations
Let the solution curve \(y=y(x)\) of the differential equation \(\left(4+x^{2}\right) d y-2 x\left(x^{2}+3 y+4\right) d x=0\) pass through the origin. Then \(y(2)\) is equal to
- A \(10\)
- B \(11\)
- C \(12\)
- D \(13\)
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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