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
- A \(4\)
- B \(3\)
- C \(1\)
- D \(2\)
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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