MCQ
વિકલ સમીકરણ $\frac{{{d^2}y}}{{d{x^2}}} - \sqrt {\frac{{dy}}{{dx}} - 3} = x$ ના કક્ષા મેળવો.
- ✓$2$
- B$1$
- C$1/2$
- D$3$
Squaring both sides, we get ${\left( {\frac{{{d^2}y}}{{d{x^2}}} - x} \right)^2} = \left( {\frac{{dy}}{{dx}} - 3} \right)$
==>${\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} + {x^2} - 2x\frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy}}{{dx}} - 3$. Clearly, degree $= 2.$
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કારણ :વિધાન $- II : x^{1/x}$ એ $0 < x < e $ માટે વધે અને $x > e $ માટે ઘટે છે.