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Vector Algebra

Question
CBSEENMA12032952

Form the differential equation of the family of curves by eliminating arbitrary constants a and b.
y = a e3x + be-2x

Solution
The equation of family of curves is
y = a e3x + be-2x                       ...(1)
Differentiating both sides w.r.t.x, we get
                 straight y subscript 1 space equals 3 space straight a space straight e to the power of 3 straight x end exponent space minus space 2 space straight b space straight e to the power of negative 2 straight x end exponent                ...(2)
Again differentiating w.r.t. x, we get, 
                 straight y subscript 2 space equals space 9 space straight a space straight e to the power of 3 straight x end exponent plus 4 space be to the power of negative 2 straight x end exponent                  ...(3)
Mutliplying (2) by 3 subtracting from (3), we get, 
                      straight y subscript 2 minus 3 straight y subscript 1 space equals space 10 space straight b space straight e to the power of negative 2 straight x end exponent
therefore space space space space straight b space straight e to the power of negative 2 straight x end exponent space equals space 1 over 10 left parenthesis straight y subscript 2 minus 3 straight y subscript 1 right parenthesis                            ...(4)
Multiplying (2) by 2 and adding it to (3), we get,
                straight y subscript 2 plus 2 straight y subscript 1 space equals space 15 space straight a space straight e to the power of 3 straight x end exponent
therefore space space space space space space space space ae to the power of 3 straight x end exponent space equals space 1 over 15 left parenthesis straight y subscript 2 plus 2 straight y subscript 1 right parenthesis                            ...(5)
From (1), (4) and (5), we get,
                               straight y space equals space 1 over 15 left parenthesis straight y subscript 2 plus 2 straight y subscript 1 right parenthesis+1 over 10 left parenthesis straight y subscript 2 minus 3 straight y subscript 1 right parenthesis
or                         straight y equals 1 over 15 straight y subscript 2 plus 2 over 15 straight y subscript 1 plus 1 over 10 straight y subscript 2 minus 3 over 10 straight y subscript 1
or                 open parentheses 1 over 15 plus 1 over 10 close parentheses space straight y subscript 2 space plus space open parentheses 2 over 15 minus 3 over 10 close parentheses space straight y subscript 1 minus straight y space equals space 0
or          open parentheses fraction numerator 2 plus 3 over denominator 30 end fraction close parentheses space straight y subscript 2 plus space open parentheses fraction numerator 4 minus 9 over denominator 30 end fraction close parentheses space straight y subscript 1 space minus space straight y space equals space 0
or      1 over 6 straight y subscript 2 minus 1 over 6 straight y subscript 1 minus straight y space equals space 0
or            straight y subscript 2 minus straight y subscript 1 minus 6 straight y equals 0 comma space which is required differential equation.n