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]
2. solve (dy/dx) = sqrt[4-(y^2)]
3. solve (dy/dx)+y=1
4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0
5.solve ( e^x + e^( –x) ) dy – ( e^x – e^(–x) ) dx = 0
6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )
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