MCQ
The order of the differential equation whose general solution is given by $y = ({c_1} + {c_2})$ $\cos (x + {c_3}) - {c_4}{e^{x + {c_5}}},$ where ${c_1},\;{c_2},\;{c_3},\;{c_4},\;{c_5}$ are arbitrary constants, is
- A$5$
- B$4$
- ✓$3$
- D$2$
✓
Answer
Correct option: C.
$3$
c
(c) $y = ({c_1} + {c_2})\cos (x + {c_3}) - {c_4}{e^{x + {c_5}}}$
${y_1} = - ({c_1} + {c_2})\sin (x + {c_3}) - {c_4}{e^{x + {c_5}}}$
${y_2} = - ({c_1} + {c_2})\cos (x + {c_3}) - {c_4}{e^{x + {C_5}}} = - y - 2{c_4}{e^{x + {c_5}}}$
==> ${y_3} = - {y_1} - 2{c_4}{e^{x + {c_5}}} = - {y_1} + {y_2} + y$
Hence the differential equation is ${y_3} - {y_2} + {y_1} - y = 0$, which is of order $3$.
(c) $y = ({c_1} + {c_2})\cos (x + {c_3}) - {c_4}{e^{x + {c_5}}}$
${y_1} = - ({c_1} + {c_2})\sin (x + {c_3}) - {c_4}{e^{x + {c_5}}}$
${y_2} = - ({c_1} + {c_2})\cos (x + {c_3}) - {c_4}{e^{x + {C_5}}} = - y - 2{c_4}{e^{x + {c_5}}}$
==> ${y_3} = - {y_1} - 2{c_4}{e^{x + {c_5}}} = - {y_1} + {y_2} + y$
Hence the differential equation is ${y_3} - {y_2} + {y_1} - y = 0$, which is of order $3$.
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