JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2} y-y-4 x^{3}\right)=0, x>1\) with \(y\,(2)=-2\). Then \(y\,(3)\) is equal to
- A \(-18\)
- B \(-12\)
- C \(-6\)
- D \(-3\)
Answer & Solution
Correct Answer
(A) \(-18\)
Step-by-step Solution
Detailed explanation
\(x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2} y-y-4 x^{3}\right)=0\) \(x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2}-1\right) y=4 x^{3}\) \(\frac{d y}{d x}+\frac{\left(3 x^{2}-1\right)}{\left(x-x^{3}\right)} y=\frac{4 x^{3}}{\left(x-x^{3}\right)}\)…
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