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 July 2020

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

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main July 2020

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

₹ 12999/- ₹ 6999/-
Buy Now
Test Series NEET July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout NEET May 2021

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

₹ 22999/- ₹ 11999/-
Buy Now
Exams
Articles
Questions