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 .