Differential Equations 1 Question 10
10. 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
(a) 5
(b) 4
(c) 3
(d) 2
(1998, 2M)
Objective Questions II
(One or more than one correct option)
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Answer:
Correct Answer: 10. (b, c)
Solution:
- Given, $y=\left(c _1+c _2\right) \cos \left(x+c _3\right)-c _4 e^{x+c _5}$
$\Rightarrow \quad y=\left(c _1+c _2\right) \cos \left(x+c _3\right)-c _4 e^{x} \cdot e^{c _5}$
Now, let $\quad c _1+c _2=A, c _3=B, c _4 e^{c _5}=c$
$\Rightarrow \quad y=A \cos (x+B)-c e^{x}$
On differentiating w.r.t. $x$, we get
$$ \frac{d y}{d x}=-A \sin (A+B)-c e^{x} $$
Again, on differentiating w.r.t. $x$, we get
$$ \begin{aligned} \frac{d^{2} y}{d x^{2}} & =-A \cos (x+B)-c e^{x} \\ \Rightarrow \quad \frac{d^{2} y}{d x^{2}} & =-y-2 c e^{x} \\ \Rightarrow \quad \frac{d^{2} y}{d x^{2}}+y & =-2 c e^{x} \end{aligned} $$
Again, on differentiating w.r.t. $x$, we get
$$ \begin{aligned} & \frac{d^{3} y}{d x^{3}}+\frac{d y}{d x}=-2 c e^{x} \\ \Rightarrow \quad & \frac{d^{3} y}{d x^{2}}+\frac{d y}{d x}=\frac{d^{2} y}{d x^{2}}+y \end{aligned} $$
[from Eq. (v)]
which is a differential equation of order 3 .
