Question

Find the general solution of the differential equation? y dy/dx-2e^x=0

Answer 1

`y = pm 2sqrt(e^x + A)`

Explanation:

We wish to solve the differential equation `y * y' - 2e^x = 0`.

This is a separable, first-order ordinary differential equation. As such, it can be solved using techniques suitable for separable 1st-order ODE's.

The most straightforward technique is to get our equation in the form `f(y) dy = g(x) dx`. We can then integrate both sides, ridding ourselves of the `y'` term.

We get our function into this form as such:

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

See that our left-hand side is a function of just `y` and our right-hand side a function of just `x`. Integrate both sides.

`int (y) dy = int (2e^x)dx`
`1/2 y^2 = 2e^x + C`
`y^2 = 4e^x + 2C`
`y = pm sqrt(4e^x + 2C) = pm 2sqrt(e^x + C/2)`

Since `C` is an arbitrary constant, let `A = C/2`. Then our final answer is `y = pm 2sqrt(e^x + A)`.