Discussion Forum

The solution  of   $\frac{d^{2}x}{dy^{2}}-x=k$ , where k is a non-zero constant , vanishes  when y=0 and tends  of finite limit as y  tends to infinity , is 

Answer:

Explanation:

 We can  write given different equation as 

    (D²-1)x= k  .....(i) 

where , $D=\frac{d}{dy}$

 Its auxiliary equation  is m²-1=0 , so that  m=1,-1

 Hence, CF= C₁e^y  +C₂ e^-y

 Where C₁ , C₂ are arbitary constants

  Now, also   $PI=\frac{1}{D^{2}-1}k$

    $=k.\frac{1}{D^{2}-1}e^{0.y}$

$=K.\frac{1}{0^{2}-1}e^{0.y}=-K$

 So,  solution  of eq.(i) is 

$x=C_{1}e^{y}+C_{2}e^{-y}-k$  .....(ii)

 Given that x=0 , when y=0

 So, 0= $C_{1}+C_{2}-k$     (From (ii))

 $\Rightarrow$   $C_{1}+C_{2}=k$

Multiplying  both sides of eq.(ii) by e^-y . we get

$x.e^{-y}=C_{1}+C_{2}e^{-2y}-ke^{-y}$ .....(iv)

 Given that x → m when  y→ $\infty$ , m being a finite quanity

 So, eq (iv) becomes

 $x\times0=C_{1}+C_{2}\times0-(k\times0)$

$\Rightarrow$   C₁=0       .........(v)

 From  eqs. (iv) and (v) , we get

 C₁  =0  and C₂ =k

 Hence , eq.(ii) becomes

$x=ke^{-y}-k=k(e^{-y}-1)$