Question

How to you find the general solution of `4yy'-3e^x=0`?

Answer 1

`y = sqrt(3/2e^x + C)`

Explanation:

Changing the notation from Lagrange's notation to Leibniz's notation we have:

`4ydy/dx-3e^x = 0`
`:. 4ydy/dx = 3e^x`

This is a First Order sepaable DE, and "seperating the variables" give is

`:. int 4y dy = int3e^x dx`

Integrating gives us:

`2y^2 = 3e^x + C_1`
`:. y^2 = 3/2e^x + C` (by writing `C_1=1/2C`)
`:. y = sqrt(3/2e^x + C)`