Differential Equations 1 Question 17
17. If $P(1)=0$ and $\frac{d P(x)}{d x}>P(x), \forall x \geq 1$, then prove that $P(x)>0, \forall x>1$.
(2003, 4M)
Show Answer
Answer:
Correct Answer: 17. (d)
Solution:
- Given, $P(1)=0$ and $\frac{d P(x)}{d x}-P(x)>0, \forall x \geq 1$
On multiplying Eq. (i) by $e^{-x}$, we get
$$ \begin{array}{cc} & e^{-x} \cdot \frac{d}{d x} P(x) \cdot \frac{d}{d x} e^{-x}>0 \\ \Rightarrow & \frac{d}{d x}\left(P(x) \cdot e^{-x}\right)>0 \\ \Rightarrow & P(x) \cdot e^{-x} \text { is an increasing function. } \\ \Rightarrow & P(x) \cdot e^{-x}>P(1) \cdot e^{-1}, \forall x \geq 1 \\ \Rightarrow & P(x)>0, \forall x>1 \quad\left[\because P(1)=0 \text { and } e^{-x}>0\right] \end{array} $$
