JEE Mains · Maths · STD 12 - 9. differential equations
The curve satisfying the differential equation, \(ydx-(x + 3y^2 )\, dy = 0\) and passing through the point \((1, 1)\) , also passes through the point
- A \(\left( {\frac{1}{4}, - \frac{1}{2}} \right)\)
- B \(\left( { - \frac{1}{3},\frac{1}{3}} \right)\)
- C \(\left( { \frac{1}{3},-\frac{1}{3}} \right)\)
- D \(\left( {\frac{1}{4}, \frac{1}{2}} \right)\)
Answer & Solution
Correct Answer
(B) \(\left( { - \frac{1}{3},\frac{1}{3}} \right)\)
Step-by-step Solution
Detailed explanation
\(y d x-x d y-3 y^{2} d y=0\) \(\frac{d x}{d y}=\frac{x}{y}+3 y\) \(\frac{d x}{d y}-\frac{x}{y}=3 y\) If \(=e^{-\int \frac{1}{y} d y}=e^{-\ln y}=\frac{1}{y}\) \(\therefore \quad\) solution is \(\frac{x}{y}=\int 3 y \cdot \frac{1}{y} d y\) \(\Rightarrow \quad \frac{x}{y}=3 y+c\)…
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