Question

Y=ae^3x+be^2x find differential equation

Answer 1

y=ae2x+be3xy=ae2x+be3x -- ( 1 ) Differentiating,y′=2ae2x+3be3xy′=2ae2x+3be3x -- ( 2 ) multiply (1) by 2 and subtract from (2) y′−2y=be3xy′−2y=be3x -- ( 3 ) Differentiating ( 3 ), y′′−2y′=3be3xy″−2y′=3be3x -- ( 4)

multiply (3) by 3 and subtract from (4) y′′−5y′+6y=0

Answer 2

The solution is on sheet .
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