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JEE Mains · Maths · STD 12 - 9. differential equations

The solution of the differential equation \(ydx - \left( {x + 2{y^2}} \right)dy = 0\) is \(x\, = f(y)\). If \(f(-1)\, = 1\), then \(f(1)\) is equal to

  1. A \(4\)
  2. B \(3\)
  3. C \(1\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

Given differential equation is \(y d x-\left(x+2 y^{2}\right) d y=0\) \(\Rightarrow \mathrm{ydx}-\mathrm{xdy}-2 \mathrm{y}^{2} \mathrm{dy}=0\) \(\Rightarrow \begin{array}{c}{y d x-x d y} \\ {y^{2}}\end{array}=2 d y\)…
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