Question

If y = P e^ax + Q e^bx, show that

`(d^y)/(dx^2)=(a+b)dy/dx+aby=0`

Answer 1

y = P e^ax + Q e^bx

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.