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.

 

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Buy Now
Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions