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An equation relating to the stability of an aeroplane is given by ` (dv)/(dt) = g cos alpha - kv`, where ` v` is the velocity and ` g, alpha, k` are constants. Find an expression for the velocity if `v = 0` at ` t =0` |
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Answer» The given differential equation is ` (dv)/(dt) + kv= g cos alpha " "` ... (i) This is of the form ` (dv)/(dt) + Pv = Q`, where `P = k and Q = g cos alpha `. Thus, the given equation is linear. `IF = e ^(int Pdx) = e ^(int k dt ) = e^(kt )` So, the solution of the given differential equation is `vxx IF = int {Q xx IF}dt + C`, i.e., `ve^(kt) = int (gcos alpha ) e^(kt) dt + C` `" " = ((g cos alpha ) e^(kt))/( k ) +C" " `... (ii) Now, it is given that ` v =0` when `t = 0` Putting `t=0 and v=0 `in (ii), we get `C = (-g cos alpha)/( k)` ` therefore ve^(kt) = ((g cos alpha ) e^(kt))/(k) - (gcos alpha )/( k)` `rArr v = (1)/(k) (g cos alpha )( 1 - e^(kt))`, which is the required expression. SOLUTION OF `(dx)/(dy) + Px =Q` Working Rule for Solving ` (dx)/(dy) +Px = Q ` (i) Find If = ` e ^(int Pdy)` (ii) The solution is given by ` x xx IF = int {Q xx IF}dy +C`. |
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