MCQ
$y + {x^2} = \frac{{dy}}{{dx}}$ has the solution
- ✓$y + {x^2} + 2x + 2 = c{e^x}$
- B$y + x + {x^2} + 2 = c{e^{2x}}$
- C$y + x + 2{x^2} + 2 = c{e^x}$
- D${y^2} + x + {x^2} + 2 = c{e^x}$
This is the linear differential equation in $y$, where $P = - 1,\,Q = {x^2}$
$I.F.$ $ = {e^{\int {P.dx} }}$$ = {e^{\int { - dx} }} = {e^{ - x}}$
Hence solution, $y.\,({\rm{I}}{\rm{.F}}). = \int {Q.({\rm{I}}{\rm{.F}})\,dx + c} $
==> $y{e^{ - x}} = - {x^2}{e^{ - x}} - 2x{e^{ - x}} - 2{e^{ - x}} + c$
==> $y + {x^2} + 2x + 2 = c{e^x}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2} .$
Match the statements in Column $I$ with the statements in Column $II$ and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the $ORS$.
| Column $I$ | Column $II$ |
| $(A)$ If $a=1$ and $b=0$, then ( $x, y$ ) | $(p)$ lies on the circle $x^2+y^2=1$ |
| $(B)$ If $a=1$ and $b=1$, then $(x, y)$ | $(q)$ lies on $\left(x^2-1\right)\left(y^2-1\right)=0$ |
| $(C)$ If $a=1$ and $b=2$, then ( $x, y)$ | $(r)$ lies on $y=x$ |
| $(D)$ If $a=2$ and $b=2$, then $(x, y)$ | $(s)$ lies on $\left(4 x^2-1\right)\left(y^2-1\right)=0$ |