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

Let `w(x)= u(x) - v(x) ...(i)`
and `h(x) = f(x) -g(x)`
On differentiating Eq. (i) w.r.t.x
`(dw)/dx= (du)/dx-(dv)/dx`
`={f(x)-p(x)cdotu(x)}-{g(x)-p(x) v(x)}` [given]
`={f(x)-g(x)}-p(x)[u(x)-u(x)]`
`rArr (dw)/dx=h(x)-p(x)cdot w(x) …(ii)`
`rArr (dw)/dx+p(x) w(x)=h(x)` shich is linear differential
equation.
The integrating factor is given by
`IF=e^(int p(x)dx)=r(x)` [let]
on multiplying both sides of Eq. (ii) of r(x), We get
`r(x)cdot (dw)/dx+p(x) (r(x)) w(x) = r(x) cdot h (x)`
`rArr d/dx[r(x) w(x)]=r(x) cdot h(x) [therefore (dr)/dx=p(x)cdotr(x)]`
Now, `r(x) =e^(intp(x)dx)gt0,AAx`
and `h(x)=f (x) -g(x)gt0,for xgtx_(1)`
Thus, `d/dx[r(x)w(x)]gt0,AAxgt x_(1)`
` r(x)w(x)` increases on the interval ` [x, infty[`
Therefore, for all `x gt x_(1)`
`r (x) w(x) gt r(x_1)w(x_1)gt0`
`[therefore r(x_1)gt0 and u (x_1) gt v(x_1)]`
`rArr w(x) gt 0AA x gt x_(1)`
`rArr u(x)gtv(x)AAxgtx_(1) [therefore r(x)gt0]`
Hence, there cannot exist a point (x,y) such that `xgtx_(1)`
and y=u (x) and y=v (x).