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)
Show Answer
Answer:
Correct Answer: 10. ( c)
Solution:
- Given, $y=\left(c _1+c _2\right) \cos \left(x+c _3\right)-c _4 e^{x+c _5}$ $\quad$ …….(i)
$\Rightarrow \quad y=\left(c _1+c _2\right) \cos \left(x+c _3\right)-c _4 e^{x} \cdot e^{c _5}$
Now, let $\quad c _1+c _2=A, c _3=B, c _4 e^{c _5}=c$
$\Rightarrow \quad y=A \cos (x+B)-c e^{x}$ $\quad$ …….(ii)
On differentiating w.r.t. $x$, we get
$ \frac{d y}{d x}=-A \sin (A+B)-c e^{x} $ $\quad$ …….(iii)
Again, on differentiating w.r.t. $x$, we get
$\frac{d^{2} y}{d x^{2}} =-A \cos (x+B)-c e^{x} $ $\quad$ …….(iv)
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}} =-y-2 c e^{x}$ $\quad$ …….(v)
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}+y =-2 c e^{x} $
Again, on differentiating w.r.t. $x$, we get
$\frac{d^{3} y}{d x^{3}}+\frac{d y}{d x}=-2 c e^{x} $ $\quad$ …….(vi)
$\Rightarrow \quad \frac{d^{3} y}{d x^{2}}+\frac{d y}{d x}=\frac{d^{2} y}{d x^{2}}+y$
[from Eq. (v)]
which is a differential equation of order 3 .
