InterviewSolution
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InterviewSolution
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Let `u(x)`and `v(x)`satisfy thedifferential equation `(d u)/(dx)+p(x)u=f(x)`and `(d v)/(dx)+p(x)v=g(x)`arecontinuous functions. If `u(x_1)`for some `x_1`and `f(x)>g(x)`for all `x > x_1,`prove thatany point `(x , y),`where `x > x_1,`does notsatisfy the equations `y=u(x)`and `y=v(x)dot` |
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Answer» 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). |
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