JEE Mains · Maths · STD 12 - 9. differential equations
If \(y (x)\) is the solution of the differential equation \(\frac{{dy}}{{dx}} + \left( {\frac{{2x + 1}}{x}} \right)y = {e^{ - 2x}},x > 0\) where \(y\,\,(1)\, = \,\frac{1}{2}{e^{ - 2}},\) then
- A \(y\,\,({\log _e}\,2)\, = \,{\log _e}\,4\)
- B \(y\,\,({\log _e}\,2)\, = \,\frac{{{{\log }_e}\,2}}{4}\)
- C \(y(x)\) is decreasing in \(\left( {\frac{1}{2},1} \right)\)
- D \(y(x)\) is decreasing in \((0, 1)\)
Answer & Solution
Correct Answer
(C) \(y(x)\) is decreasing in \(\left( {\frac{1}{2},1} \right)\)
Step-by-step Solution
Detailed explanation
I.F. \(=e^{\int\left(2+\frac{1}{x}\right) d x}=e^{2 x} \cdot x\) Solution will be \(y\left(x e^{2 x}\right)=\int e^{-2 x} \cdot x e^{2 x} \cdot d x+c\) \(x y e^{2 x}=\frac{x^{2}}{2}+c\)…
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