Answer 1

y = (c₁ + c₂ x) e^x

∴ ye ^-x = c₁ + c₂x

Differentiating w.r.t. x, we get

`y (-e^-x) + e ^-x  dy/dx = 0 + c_2`

∴`e^-x(dy/dx - y) = c_2`

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

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

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

∴`(d^2y)/dx^2 - 2dy/dx + y = 0`