JEE Mains · Maths · STD 12 - 9. differential equations
If the solution curve of the differential equation \(\left(2 x-10 y^{3}\right) d y+y d x=0\), passes through the points \((0,1)\) and \((2, \beta)\), then \(\beta\) is a root of the equation:
- A \(y^{5}-2 y-2=0\)
- B \(2 y^{5}-2 y-1=0\)
- C \(2 y^{5}-y^{2}-2=0\)
- D \(y^{5}-y^{2}-1=0\)
Answer & Solution
Correct Answer
(D) \(y^{5}-y^{2}-1=0\)
Step-by-step Solution
Detailed explanation
\(\left(2 x-10 y^{3}\right) d y+y d x=0\) \(\Rightarrow \frac{d x}{d y}+\left(\frac{2}{y}\right) x=10 y^{2}\) \(\text { I. } F .=e^{\int \frac{2}{y} d y}=e^{2 \ell n(y)}=y^{2}\) Solution of \(D.E.\) is \(\therefore \quad x \cdot y=\int\left(10 y^{2}\right) y^{2} \cdot d y\)…
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