Find the differential equation of the family of curves y=Ae^{2x}+Be^{-2x}, where A and B are arbitrary constants.

 

 

 

 

 
 
 
 
 

Answers (1)

y=Ae^{2x}+Be^{-2x},  [Given] ----(1)

On differentiating equation (1) w.r.t x, we get

\frac{\mathrm{d} y}{\mathrm{d} x}=2\cdot Ae^{2x}-2Be^{2x}-----(ii)

Again, differentiating equation (ii) w.r.t x, we get

\frac{\mathrm{d} ^2y}{\mathrm{d} x^2}=4Ae^{2x}-4Be^{-2x}

         =4(Ae^{2x}+Be^{-2x})

        \Rightarrow 4y

\therefore \frac{\mathrm{d} ^2y}{\mathrm{d} x^2}-4y=0 is the required differential equation.

 

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