solve ( e^x + e^( –x) ) dy – ( e^x – e^(–x) ) dx = 0

 

ncert cbse mathematics  class 12th differential equations

exercise 9.4 variable separable differential equations type

 

5.solve  ( e^x + e^( –x)  ) dy – ( e^x – e^(–x)  ) dx = 0

This is a differential equation of the type variable separable.

( e^x + e^( –x)  ) dy – ( e^x – e^(–x)  ) dx = 0

dy ={ [ e^x – e^(–x)  ]  / [e^x + e^( –x)] } dx

now integrate both sides

keeping in mind that integral of {(f ') / f }form is ln(f)  

{ here ln refers to natural logarithms }

 

y=ln{[e^x + e^( –x)]} +C

6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )

This is a differential equation of the type variable separable.

dy / [  1 + (y^2) ] = [ 1+( x^2) ]dx

now integrating on both sides using integration formula

arc[tan y] =x +[(x^3)/3] +C

note that arc[tan y] stands for  inverse  tan  of (y)

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 ncert cbse 12th mathematics

exercise 9.4

differential equations 

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

3. solve (dy/dx)+y=1

solution

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

 solution

   

5.solve  ( e^x + e^( –x)  ) dy – ( e^x – e^(–x)  ) dx = 0

solution

6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )

solution