JEE Advance - Mathematics (1997 - No. 2)
Let $$u(x)$$ and $$v(x)$$ satisfy the differential equation $${{du} \over {dx}} + p\left( x \right)u = f\left( x \right)$$ and $${{dv} \over {dx}} + p\left( x \right)v = g\left( x \right),$$ 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 $$y=v(x)$$
If u(x_1) > v(x_1) and f(x) > g(x) for all x > x_1, then u(x) > v(x) for all x > x_1.
Subtracting the two differential equations, we get d(u-v)/dx + p(x)(u-v) = f(x) - g(x).
Multiplying by the integrating factor exp(∫p(x)dx), we get d/dx [exp(∫p(x)dx) (u-v)] = exp(∫p(x)dx) (f(x) - g(x)).
Integrating from x_1 to x, we get exp(∫p(t)dt from x_1 to x) (u(x) - v(x)) - (u(x_1) - v(x_1)) = ∫ exp(∫p(t)dt) (f(t) - g(t)) dt from x_1 to x.
Since u(x_1) > v(x_1) and f(x) > g(x) for x > x_1, it follows that u(x) > v(x) for all x > x_1.
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