Question

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

y = c₁e ^3x + c₂e ^2x

Answer 1

y = c₁e ^3x + c₂e ^2x

Dividing throughout by e^2x , we get

ye ^-2x = c₁e ^x + c₂

Differentiating w.r.t. x, we get

`-2ye^ -2x + e^-2x dy/dx = c_1e^ x`

∴ `e^-2x(dy/dx-2y) = c_1e^x`

Dividing throughout by e^x , we get

`e^-3x(dy/dx - 2y) = c_1`

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

`e^-3x((d^2y)/dx^2 - 2 dy/dx) - 3e^(-3x)(dy/dx- 2y ) = 0`

∴ `e^-(3x)((d^2y)/dx^2 - 2 dy/dx-3 dy/dx+ 6y ) = 0`

∴ `(d^2y)/dx^2- 5dy/dx + 6y = 0`