JEE Mains · Maths · STD 12 - 9. differential equations
If the solution curve \( y=f(x) \) of the differential equation \( (x^{2}-4)y^{\prime}-2xy+2x(4-x^{2})^{2}=0, x>2 \) passes through the point (3, 15), then the local maximum value of f is:
- A 16
- B 12
- C 8
- D 20
Answer & Solution
Correct Answer
(A) 16
Step-by-step Solution
Detailed explanation
\(\left(x^2-4\right) y^1-2 x y=-2 x\left(x^2-4\right)^2\) \(d\left(\frac{y}{x^2-4}\right)=-2 x\) \(y=\left(-x^2+C\right)\left(x^2-4\right)\) for \(x=3 \quad y=15 \Rightarrow C=12\) \(y =\left(- x ^2+12\right)\left( x ^2-4\right)\) \(y^1=0 \Rightarrow x=2 \sqrt{2}\)…
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