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Question
If y = P eax + Q ebx, show that
`(d^y)/(dx^2)=(a+b)dy/dx+aby=0`
Solution
y = P eax + Q ebx
Differentiating w.r.t x, we get:
`dy/dx=Pae^(ax)+Qbe^(bx)..................(1)`
`(a+b)dy/dx=(a+b)(Pae^(ax)+Qbe^(bx))`
`(a+b)dy/dx=Pa^2e^(ax)+Qb^2e^(bx)+ab(Pe^(ax)+Qe^(bx))`
`(a+b)dy/dx=Pa^2e^(ax)+Qb^2e^(bx)+aby`
`-[-(a+b)dy/dx+aby]=Pa^2e^(ax)+Qb^2e^(bx)........(2)`
Differentiating (1) w.r.t. x, we get:
`(d^y)/(dx^2)=Pa^2e^(ax)+Qb^2e^(bx)................(3)`
Subtracting (2) from (3), we get:
`(d^y)/(dx^2)-(a+b)dy/dx+aby=Pa^2e^(ax)+Qb^2e^(bx)-Pa^2e^(ax)-Qb^2e^(bx)`
`(d^y)/(dx^2)-(a+b)dy/dx+aby=0`
Hence proved.
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