JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(\frac{ dy }{ dx }=\frac{4 y ^{3}+2 yx ^{2}}{3 xy ^{2}+ x ^{3}}, y (1)=1\). If for some \(n \in N , y (2) \in[ n -1, n )\), then \(n\) is equal to \(\dots\dots\)
- A \(2\)
- B \(3\)
- C \(4\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=\frac{y}{x} \frac{\left(4 y^{2}+2 x^{2}\right)}{\left(3 y^{2}+x^{2}\right)}\) Put \(y=v x \quad \Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x}\) \(v+x \frac{d v}{d x}=\frac{v\left(4 v^{2}+2\right)}{\left(3 v^{2}+1\right)}\)…
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