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${y^2} + x + {x^2} + 2 = c{e^x}$
✓
Answer
Correct option: A.
$y + {x^2} + 2x + 2 = c{e^x}$
a
(a) $y + {x^2} = \frac{{dy}}{{dx}}$ ==> $\frac{{dy}}{{dx}} - y = {x^2}$
(a) $y + {x^2} = \frac{{dy}}{{dx}}$ ==> $\frac{{dy}}{{dx}} - y = {x^2}$
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} $
==> $y{e^{ - x}} = - {x^2}{e^{ - x}} - 2x{e^{ - x}} - 2{e^{ - x}} + c$
==> $y + {x^2} + 2x + 2 = c{e^x}$.
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