The differential equation whose general solution is given by,
$y = \left( {{c_1}\cos (x + {c_2})} \right) - ({c_3}{e^{( - x + {c_4})}}) + ({c_5}\sin x)$ , where $c_1, c_2, c_3, c_4, c_5$ are arbitrary constants, is
- A
$\frac{{{d^4}y}}{{d{x^4}}}\,\, - \,\,\frac{{{d^2}y}}{{d{x^2}}}\,\, + \,y\, = \,0\,$
- ✓
$\frac{{{d^3}y}}{{d{x^3}}}\,\, + \,\,\frac{{{d^2}y}}{{d{x^2}}}\,\, + \,\,\frac{{dy}}{{dx}}\, + \,y\, = \,0\,$
- C
$\frac{{{d^5}y}}{{d{x^5}}}\,\, + \,y\, = \,0\,$
- D
$\frac{{{d^3}y}}{{d{x^3}}}\,\, - \,\,\frac{{{d^2}y}}{{d{x^2}}}\,\, + \,\,\frac{{dy}}{{dx}}\, - \,y\, = \,0\,$
✓
Answer
Correct option: B.$\frac{{{d^3}y}}{{d{x^3}}}\,\, + \,\,\frac{{{d^2}y}}{{d{x^2}}}\,\, + \,\,\frac{{dy}}{{dx}}\, + \,y\, = \,0\,$
b
$y = c_1 \cos(x + c_2) - (c_3 {e^{ - \,x\,\, + \,\,{c_4}}} ) + (c_5 \sin x)$
$y = c_1 (\cos x \cos c_2 - \sin x \sin c_2) - (c_3 {e^{\,{c_4}}} \, e^{-x})+ (c_5 \sin x)$
$y = (c_1 \cos c_2 ) \cos x - (c_1 \sin c_2 -c_5) \sin x - (c_3 ) e^{-x}$
$y = l \cos x + m \sin x - n e^{ -x}....(i)$
where $l, m, n$ are arbitrary constant
$\frac{{dy}}{{dx}}\, = - l \sin x + m \cos x + n e^{-x}....(ii)$
$\frac{{d^2y}}{{dx^2}}\,= - l \cos x - m \sin x - n e^{-x}....(iii)$
$\frac{{d^3y}}{{dx^3}}\, = l \sin x - m \cos x + n e^{-x}....(iv)$
$(i) + (iii)$ gives $\frac{{{d^2}y}}{{d{x^2}}}\, + y = - 2n e^{-x}....(v)$
$(ii) + (iv)$ gives $\frac{{{d^3}y}}{{d{x^3}}}\, + \,\frac{{dy}}{{dx}}\,= 2n e^{-x} ....(vi)$
From $(v)$ and $(vi)$ we get $\frac{{{d^3}y}}{{d{x^3}}}\, + \,\frac{{dy}}{{dx}}\, = \, - \,\left( {\frac{{{d^2}y}}{{d{x^2}}}\, + \,y} \right)$
or $\frac{{{d^3}y}}{{d{x^3}}}\,\, + \,\,\frac{{{d^2}y}}{{d{x^2}}}\,\, + \,\,\frac{{dy}}{{dx}}\, + \,y\, = \,0\,$ is the required differential equation
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