Question

How can i solve this equation ?

y'-y=e^2x using linear defferential equation

Answer 1

`y = e^x (e^x + C)`

Explanation:

Simplest approach is to use an Integrating factor.

For:

• `y' + f(x) y = g(x)`,

there may be a function:

• `bb I(x) = exp (int f(x) dx)`,

which when applied to the LHS makes it exact.

Ie if we can find `bb I (x)`, then:

• `bb I(x) y' + bb I(x) f(x) y = (bb I(x) y)^'`

So here `bb I = e^(int (-1) dx) = e^(-x + C) = Ce^(-x)`

It is usual to drop the constant `C`, as it cancels out on both sides, so we have:

`y' - y = e^(2x)`

`e^(-x)y' - e^(-x)y = e^(-x)* e^(2x)`

`(e^(-x)y)^' = e^(x)`

`e^(-x)y = e^(x) + C`

`y = e^(2x) + Ce^x = e^x (e^x + C)`