Let u(x) and v(x) satisfy the...

MathematicsLinear Differential EquationsJEE Advanced 1997Moderate

Let $u (x)$ and $v (x)$ satisfy the differential equation $\frac{d u}{d x} + p (x) u = f (x)$ and $\frac{d v}{d x} + p (x) v = g (x)$ , where $p (x), f (x)$ and $g (x)$ are continuous functions. If $u (x_{1}) > v (x_{1})$ 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 $y = v (x)$ .

Visualized Solution (Hindi)

Visualizing the Problem

Given differential equations:
1. $\frac{d u}{d x} + p (x) u = f (x)$
2. $\frac{d v}{d x} + p (x) v = g (x)$
Initial condition: $u (x_{1}) > v (x_{1})$
Growth condition: $f (x) > g (x)$ for all $x > x_{1}$

Defining the Difference Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w (x)

Define a new function $w (x) = u (x) - v (x)$
At $x = x_{1}$ , $w (x_{1}) = u (x_{1}) - v (x_{1}) > 0$
Our goal is to show $w (x) > 0$ for all $x > x_{1}$

Subtracting the Equations

Subtracting (2) from (1):
$(\frac{d u}{d x} - \frac{d v}{d x}) + p (x) (u - v) = f (x) - g (x)$
Substituting $w = u - v$ :
$\frac{d w}{d x} + p (x) w = f (x) - g (x)$

Applying the Integrating Factor

Integrating Factor $I . F . = e^{\int p (x) d x}$
Multiplying the DE by $I . F .$ :
$e^{\int p (x) d x} \frac{d w}{d x} + p (x) e^{\int p (x) d x} w = (f (x) - g (x)) e^{\int p (x) d x}$
This simplifies to:
$\frac{d}{d x} (w (x) e^{\int p (x) d x}) = (f (x) - g (x)) e^{\int p (x) d x}$

Analyzing the Sign of the Derivative

Since $f (x) > g (x)$ and $e^{\int p (x) d x} > 0$ :
$(f (x) - g (x)) e^{\int p (x) d x} > 0$ for all $x > x_{1}$
Therefore, $\frac{d}{d x} (w (x) e^{\int p (x) d x}) > 0$
This implies $h (x) = w (x) e^{\int p (x) d x}$ is a strictly increasing function for $x > x_{1}$

Final Conclusion

At $x = x_{1}$ , $w (x_{1}) > 0$ and $e^{\int p (x) d x} > 0$ , so $w (x_{1}) e^{\int p (x) d x} > 0$
Since the function is strictly increasing for $x > x_{1}$ :
$w (x) e^{\int p (x) d x} > w (x_{1}) e^{\int p (x) d x} > 0$
This implies $w (x) > 0$ for all $x > x_{1}$
Conclusion: $u (x) > v (x)$ , so the curves $y = u (x)$ and $y = v (x)$ never intersect for $x > x_{1}$

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Visualized Solution (Hindi)

Visualizing the Problem

Defining the Difference Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w(x)

Subtracting the Equations

Applying the Integrating Factor

Analyzing the Sign of the Derivative

Final Conclusion

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