Question

Obtain the differential equation by eliminating arbitrary constants from the following equations.

y = Ae^3x + Be^−3x

Answer 1

y = Ae^3x + B.e^−3x ......(i)

Differentiating w.r.t. x, we get

`dy/dx = 3Ae^(3x) - 3 Be ^(-3x)`

Again, differentiating w.r. t. x, we get

`(d^2y)/dx^2 = 3A d/dxe^(3x) -3Bd/dx(e^(-3x))`

= `3A(3e^(3x)) - 3b(-3e^(-3x))`

=`9Ae^(3x)+9Be^(-3x)`

=`9(Ae^(3x)+Be^(-3x)) = 9y` ......[From(i)]

∴ `(d^2y)/dx^2 = 9y`