JEE Mains · Maths · STD 12 - 9. differential equations
The slope of tangent at any point \((x, y)\) on a curve \(y = y ( x )\) is \(\frac{x^2+y^2}{2 x y},x > 0\). If \(y(2)=0\), then a value of \(y(8)\) is
- A \(-2 \sqrt{3}\)
- B \(4 \sqrt{3}\)
- C \(2 \sqrt{3}\)
- D \(-4 \sqrt{2}\)
Answer & Solution
Correct Answer
(B) \(4 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=\frac{1+\left(\frac{y}{x}\right)^2}{2\left(\frac{y}{x}\right)}\) Let \(y=t x\) \(\Rightarrow t+x \frac{d t}{d x}=\frac{1+t^2}{2 t}\) \(\Rightarrow x \frac{d t}{d x}=\frac{1-t^2}{2 t}\) \(\Rightarrow \int \frac{2 t}{1-t^2} d t=\int \frac{d x}{x}\)…
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