Differential Equations 2 Question 23
24. Let $u(x)$ and $v(x)$ satisfy the differential equations $\frac{d u}{d x}+p(x) u=f(x)$ and $\frac{d v}{d x}+p(x) v=g(x), \quad$ where $p(x), f(x)$ and $g(x)$ are continuous functions. If $u\left(x _1\right)>v\left(x _1\right)$ for some $x _1$ and $f(x)>g(x)$ for all $x>x _1$, prove that any point $(x, y)$ where $x>x _1$ does not satisfy the equations $y=u(x)$ and $y=v(x)$.
$(1997,5$ M)
Integer Answer Type Question
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Answer:
Correct Answer: 24. (0)
Solution:
- Let
$w(x)=u(x)-v(x)$
and
$$ h(x)=f(x)-g(x) $$
On differentiating Eq. (i) w.r.t. $x$
$\frac{d w}{d x}=\frac{d u}{d x}-\frac{d v}{d x}$
$={f(x)-p(x) \cdot u(x)}-{g(x)-p(x) v(x)} \quad$ [given]
$={f(x)-g(x)}-p(x)[u(x)-v(x)]$
$\Rightarrow \frac{d w}{d x}=h(x)-p(x) \cdot w(x)$
$\Rightarrow \frac{d w}{d x}+p(x) w(x)=h(x)$ which is linear differential equation
The integrating factor is given by
$$ IF=e^{\int p(x) d x}=r(x) $$
On multiplying both sides of Eq. (ii) of $r(x)$, we get
$$ \begin{aligned} & r(x) \cdot \frac{d w}{d x}+p(x)(r(x)) w(x)=r(x) \cdot h(x) \\ \Rightarrow & \frac{d}{d x}[r(x) w(x)]=r(x) \cdot h(x) \quad \because \frac{d r}{d x}=p(x) \cdot r(x) \end{aligned} $$
Now,
$$ r(x)=e^{\int P(x) d x}>0, \forall x $$
and
$$ h(x)=f(x)-g(x)>0, \text { for } x>x _1 $$
Thus,
$$ \frac{d}{d x}[r(x) w(x)]>0, \forall x>x _1 $$
$r(x) w(x)$ increases on the interval $[x, \infty$ [
Therefore, for all $x>x _1$
$$ \begin{aligned} r(x) w(x) & >r\left(x _1\right) w\left(x _1\right)>0 \\ & {\left[\because r\left(x _1\right)>0 \text { and } u\left(x _1\right)>v\left(x _1\right)\right] } \\ w(x) & >0 \forall x>x _1 \\ u(x)>v(x) \forall x>x _1 & {[\because r(x)>0] } \end{aligned} $$
$$ \begin{array}{ll} \Rightarrow & w(x)>0 \forall x>x _1 \\ \Rightarrow & u(x)>v(x) \forall x>x _1 \end{array} $$
Hence, there cannot exist a point $(x, y)$ such that $x>x _1$ and $y=u(x)$ and $y=v(x)$.
