Question

How do you solve `y'+3y=0` given y(0)=4?

Answer 1

`y = 4e^(-3x)`

Explanation:

We have `y'+3y=0`, or:

`dy/dx + 3y = 0`

This is a first order linear ordinary differential equation and so we can derive an Integrating Factor using:

IF `= e^(int 3dx)`
`\ \ \ = e^(3x)`

So multiplying the DE by the IF gives:

`e^(3x)dy/dx + 3ye^(3x) = 0`

Thanks to the product rule this can now be written as the derivative of a single product:

`d/dx(ye^(3x)) = 0`

Which can now easily be solved to give;

`ye^(3x) = A`
`:. y = Ae^(-3x)`

We know that `y(0)=4 => Ae^0=4 => A=4`

Hence, the solution is:
`y = 4e^(-3x)`

Answer 2

`y=4e^(-3x)`

Explanation:

We have the separable differential equation:

`y'+3y=0`

Which can be rearranged as:

`dy/dx+3y=0`

`dy/dx=-3y`

Rearranging this by separating the variables, that is, treating `dy/dx` as division and getting the `y` terms on one side and the `x` terms on the other, we see that:

`dy/y=-3dx`

Integrate both sides:

`intdy/y=-3intdx`

`lnabsy=-3x+C`

Use the initial condition `y(0)=4` to solve for `C`:

`lnabs4=-3(0)+C`

`C=ln(4)`

Then:

`lnabsy=-3x+ln(4)`

Solving for `y`:

`absy=e^(-3x+ln(4))`

Rewriting using exponent rules:

`absy=e^(-3x)e^ln(4)`

`absy=4e^(-3x)`

`4e^(-3x)` is positive for all real values of `x`, so the absolute value bars are not necessary.

`y=4e^(-3x)`