Let u(x) and v(x) satisfy the differential equation du/dx + p(x)u = f(x) and dv/dx + p(x)v = g(x), w

Q: Why do we subtract the equations instead of solving them individually?

Solving individual differential equations is often impossible if $p(x)$, $f(x)$, or $g(x)$ are arbitrary continuous functions. Subtraction is a powerful strategic move; it isolates the *difference* between the systems, which is exactly what the question asks about. We don't need to know the exact path of $u(x)$ or $v(x)$, only the path of their difference.

Q: Does the sign of $p(x)$ matter for the integrating factor?

That is a brilliant question! The integrating factor is $e^{\int p(x) dx}$. Since the exponential function is always positive regardless of the exponent, the integrating factor will never change the sign of our inequality. It acts as a 'scaling factor' that preserves the positivity of our difference function.

Q: Is it possible for $w(x)$ to become zero later?

If $w(x)$ were to become zero, it would contradict the fact that our function $h(x) = w(x) e^{\int p(x) dx}$ is strictly increasing and started at a positive value. A strictly increasing function that starts positive can never touch zero. Therefore, the curves are 'locked' in their relative positions forever.

Question

Let $u(x)$ and $v(x)$ satisfy the differential equation $\frac{du}{dx} + p(x)u = f(x)$ and $\frac{dv}{dx} + p(x)v = g(x)$, 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 $y = v(x)$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have two particles, $u$ and $v$, moving along the x-axis. Their paths are governed by two differential equations:

$$\frac{du}{dx} + p(x)u = f(x)$$

$$\frac{dv}{dx} + p(x)v = g(x)$$

We are given that at a specific point $x_1$, $u(x_1) > v(x_1)$. This means $u$ is physically above $v$. We aim to prove that for all $x > x_1$, $u$ will **never intersect** $v$.

Instead of solving for $u(x)$ and $v(x)$ individually, we focus on the **gap** between them. Let us define a new function, $w(x) = u(x) - v(x)$. If we can prove that $w(x) > 0$ for all $x > x_1$, we have proven that the curves never intersect.

### The Algebraic Bridge

Now, let us perform the subtraction. By subtracting the second equation from the first, we obtain:

$$\left( \frac{du}{dx} - \frac{dv}{dx} \right) + p(x)(u - v) = f(x) - g(x)$$

Substituting our definition $w(x) = u(x) - v(x)$, the equation transforms into a standard first-order linear differential equation:

$$\frac{dw}{dx} + p(x)w = f(x) - g(x)$$

This is the heart of the problem. We have reduced a complex system of two equations into a single, elegant equation that describes the evolution of the gap between the two functions.

### The Power of the Integrating Factor

To solve this, we invoke the classic tool of the Integrating Factor ($I.F.$). We multiply the entire equation by $e^{\int p(x) dx}$. This specific multiplier transforms the left-hand side into the derivative of a product.

Multiplying both sides, we get:

$$e^{\int p(x) dx} \frac{dw}{dx} + p(x) e^{\int p(x) dx} w = (f(x) - g(x)) e^{\int p(x) dx}$$

By the product rule of differentiation, the left side is exactly:

$$\frac{d}{dx} \left( w(x) e^{\int p(x) dx} \right) = (f(x) - g(x)) e^{\int p(x) dx}$$

We have expressed the rate of change of a new function, $h(x) = w(x) e^{\int p(x) dx}$, in terms of the given functions $f(x)$ and $g(x)$.

### The Climax of Monotonicity

Consider the right-hand side of our equation. We are given that $f(x) > g(x)$ for all $x > x_1$, which implies $f(x) - g(x) > 0$. Furthermore, the exponential function $e^{\int p(x) dx}$ is always strictly positive.

Therefore, the product $(f(x) - g(x)) e^{\int p(x) dx}$ must be strictly positive. This implies:

$$\frac{d}{dx} (h(x)) > 0$$

If the derivative of a function is positive, the function is strictly increasing. Since $h(x_1) = w(x_1) e^{\int p(x_1) dx}$ and we know $w(x_1) > 0$, it follows that $h(x_1) > 0$.

Because $h(x)$ starts at a positive value at $x_1$ and is strictly increasing for all $x > x_1$, it can never return to zero or become negative. Since $h(x) = w(x) e^{\int p(x) dx}$ and the exponential term is always positive, $w(x)$ must also remain strictly positive for all $x > x_1$.

### Conclusion

We have arrived at the finish line. Because $w(x) > 0$, it follows that $u(x) - v(x) > 0$, or **$u(x) > v(x)$**. The curves are locked in their relative positions, and they will never intersect.

This problem teaches us that in mathematics, we do not always need to know the exact path of every variable. By focusing on the relationship between variables—the 'gap'—we can understand the behavior of the entire system.

Visualized Solution

Visualizing the Initial Condition $u (x_{1}) > v (x_{1})$

Defining the Difference Function $w (x)$

Subtracting the Differential Equations

Identifying the Integrating Factor $I . F .$

Multiplying by $e^{\int p (x) d x}$

Condensing the Left Hand Side

Analyzing the Sign of the Derivative

Deducing the Monotonicity of $h (x)$

The Final Conclusion $u (x) > v (x)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Algebraic Bridge

The Power of the Integrating Factor

The Climax of Monotonicity

Conclusion

Similar Questions

Let $y^{'} (x) + y (x) g^{'} (x) = g (x) g^{'} (x)$ , $y (0) = 0$ , $x \in R$ , where $f^{'} (x)$ denotes $\frac{df ( x )}{d x}$ and $g (x)$ is a given non-constant differentiable function on $R$ with $g (0) = g (2) = 0$ . Then the value of $y (2)$ is

Let $y = y_{1} (x)$ and $y = y_{2} (x)$ be two distinct solutions of the differential equation $\frac{d y}{d x} = x + y$ , with $y_{1} (0) = 0$ and $y_{2} (0) = 1$ respectively. Then, the number of points of intersection of $y = y_{1} (x)$ and $y = y_{2} (x)$ is

Let $f (x), x \geq 0$ , be a non-negative continuous function, and let $F (x) = \int_{0}^{x} f (t) d t, x \geq 0$ . If for some $c > 0, f (x) \leq c F (x)$ for all $x \geq 0$ , then show that $f (x) = 0$ for all $x \geq 0$ .

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

For any real numbers $α$ and $β$ , let $y_{α, β} (x), x \in R$ , be the solution of the differential equation $\frac{d y}{d x} + α y = x e^{β x}, y (1) = 1$ . Let $S = {y_{α, β} (x) : α, β \in R}$ . Then which of the following functions belong(s) to the set $S$ ?

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{3}{2}}^{\frac{3}{2}} f (x) d x$ is equal to

Let $y = y (x)$ be the solution curve of the differential equation $sin (2 x^{2}) lo g_{e} (tan x^{2}) d y + (4 x y - 42 x sin (x^{2} - \frac{π}{4})) d x = 0, 0 < x < \frac{π}{2}$ , which passes through the point $(\frac{π}{6}, 1)$ . Then $∣ y (\frac{π}{3}) ∣$ is equal to

Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $lim_{t \to x} \frac{t ^{2} y ( x ) - x ^{2} y ( t )}{x - t} = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

Visualized Solution

Visualizing the Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u(x1​)>v(x1​)

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