Q86 / 200 · Manipal (MET) 2012 · Mathematics
The order of the differential equation whose general solution is given by
$y=\left(C_1+C_2\right) \cos \left(x+C_3\right)-C_4 e^{x+C_5}$
where, $C_1, C_2, C_3, C_4, C_5$ are arbitrary constants, is
- A5
- B4
- Correct answerC3
- D2
Solution
We have,
$\begin{aligned}
y & =\left(C_1+C_2\right) \cos \left(x+C_3\right)-C_4 e^{x+C_5} \quad \mathrm{...(i)} \\
\Rightarrow y & =\left(C_1+C_2\right) \cos \left(x+C_3\right)-C_4 e^x \cdot e^{C_5}
\end{aligned}$
Now, let $C_1+C_2=A, C_3=B, C_4 e^{C_5}=C$
$\Rightarrow \quad y=A \cos (x+B)-C e^x \qquad \mathrm{...(ii)}$
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=-A \sin (x+B)-C e^x \quad \mathrm{...(iii)}$
Again, differentiating w.r.t. $x$, we get
$\begin{array}{ll}
& \frac{d^2 y}{d x^2}=-A \cos (x+B)-C e^x \quad \mathrm{...(iv)} \\
\Rightarrow \quad & \frac{d^2 y}{d x^2}=-y-2 C e^x \\
\Rightarrow \quad & \frac{d^2 y}{d x^2}+y=-2 C e^x \quad \mathrm{...(v)}
\end{array}$
Again, differentiating w.r.t. $x$, we get
$\frac{d^3 y}{d x^3}+\frac{d y}{d x}=-2 C e^x \quad \mathrm{...(vi)}$
$\Rightarrow \quad \frac{d^3 y}{d x^3}+\frac{d y}{d x}=\frac{d^2 y}{d x^2}+y[\text { from Eq. (v) }]$
which is a differential equation of order 3.
Hence, the correct answer is option: 'C'.