Let `u(x)` and `v(x)` satisfy the differential equation `(d u)/(dx)+p(x)u=f(x)` and `(d v)/(dx)+p(x)v=g(x)` are continuous functions. If `u(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)dot`

Explanation

We have following information:
a `y=u(x)` and `y=v(x)` are solutions of given differential equations.
b. `u(x_(1)) gt v(x_(1))` for some `x_(1)`
c. `f(x) gt g(x) AA x_(1)`
`(du)/(dx) +p(x)u=f(x)` and `(dv)/(dx) +p(x)v=g(x)`
or `(d(u-v))/(dx) + p(x)(u-v)+ e^(intpdx) p(x) (u-v)`
`=e^(intpdx) (f(x)-g(x))`
or `d/(dx)[(u-v).e^(intpdx)]=[f(x)-g(x)]e^(intpdx)`
Given `f(x) gt g(x) AA x gt x_(1)` and exponential function is always positive. Then RHS is positive.
`therefore d/(dx)[(u-v).e^(intpdx)] gt 0`
Hence, the function F(x) =`(u-v)e^(int(pdx))` is an increasing function.
Again, `u(x_(1)) gt v(x_(1))` for some `x_(1)`.
`therefore F(x) = (u-v)e^(intpdx)` is `+ve` at `x=x_(1)`
or `F(x) = (u-v) e^(intpdx)` is `+ve AA xgt x_(1)` `therefore` F being increasing function.
`therefore u(x) gt u(x_(1)) AA x gt x_(1)`
Hence, there is no point (x,y) such that `x gt x_(1)` which can satisfy the equations, `y=u(x)` and `y=v(x)` simultaneously.