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Question

Find the differential equation corresponding to y=ae2x+be3x+cex

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Solution

y=ae2x+be3x+cex..........(1)
Differentiating above equation w.r.t. x, we have
y=2ae2x3be3x+cex..........(2)
On subtracting eqn(1) from (2), we have
yy=2ae2x3be3x+cex(ae2x+be3x+cex)
yy=2ae2x3be3x+cexae2xbe3xcex
y'y=ae2x4be3x..........(3)
Differentiating above equation w.r.t. x, we have
y′′y=2ae2x+12be3x..........(4)
Multiplying eqn(3) by 2, we have
2y2y=2ae2x8be3x..........(5)
On subtracting eqn(5) from (4), we have
(y′′y)(2y2y)=2ae2x+12be3x(2ae2x8be3x)
y′′y2y+2y=2ae2x+12be3x2ae2x+8be3x
y′′3y'+2y=20be3x..........(6)
Differentiating above equation w.r.t. x, we have
y'''3y''+2y'=60be3x..........(7)
Multiplying eqn(6) by 3, we have
3y′′9y+6y=60be3x..........(8)
On adding eqn(7)&(8), we have
(y'''3y''+2y')+(3y′′9y+6y)=60be3x+(60be3x)
y'''3y''+2y'+3y′′9y+6y=60be3x+60be3x
y'''7y'+6y=0
Hence, the correct answer is y′′′7y+6y=0.

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