Do you dare

Let u(x) and v(x) satisfy the differential equations

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

(This question has been asked in the IIT 97 Second Exam)

#Calculus

Easy Math Editor

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Comments

Did you get it eventually? All you have to show is that the two curves cannot intersect. From the linear differential equation, and the inequality, it should be straightforward.

Ameya Daigavane - 5 years, 3 months ago

@Rishabh Cool, can you please also solve this one if you have some time at your hand.

Akhilesh Prasad - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$