`y'-y = 16` Solve the first-order differential equation

`y' - y = 16`

To solve, rewrite the derivative as `dy/dx` .

`(dy)/dx - y = 16`

Then, express the equation in the form `N(y)dy = M(x) dx` .

`(dy)/dx = y+16`

`(dy)/(y+16) = dx`

Take the integral of both sides.

`int (dy)/(y+16) = int dx`

For the left side of the equation, apply the formula `int (du)/u = ln|u|+C` .

And for the right side, apply the formula `int adx =ax + C`.

`ln |y+ 16| + C_1 = x + C_2`

Then, isolate the y. To do so, move the C1 to the right side.

`ln|y+16| = x + C_2-C_1`

Since C1 and C2 represent any number, express it as a single constant C.

`ln|y+16| = x + C`

Then, convert this to exponential equation.

`y+16=e^(x+C)`

And, move the 16 to the right side.

`y = e^(x+C) - 16`

Therefore, the general solution is `y = e^(x+C)-16` .