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

Friday, January 2, 2015

$\displaystyle{y}'-{y}={16}$

To solve, rewrite the derivative as $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}$ .

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}-{y}={16}$

Then, express the equation in the form $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ .

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}={y}+{16}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{y}+{16}}}={\left.{d}{x}\right.}$

Take the integral of both sides.

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{y}+{16}}}=\int{\left.{d}{x}\right.}$

For the left side of the equation, apply the formula $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$ .

And for the right side, apply the formula $\displaystyle\int{a}{\left.{d}{x}\right.}={a}{x}+{C}$ .

$\displaystyle{\ln}{\left|{y}+{16}\right|}+{C}_{{1}}={x}+{C}_{{2}}$

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

$\displaystyle{\ln}{\left|{y}+{16}\right|}={x}+{C}_{{2}}-{C}_{{1}}$

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

$\displaystyle{\ln}{\left|{y}+{16}\right|}={x}+{C}$

Then, convert this to exponential equation.

$\displaystyle{y}+{16}={{e}}^{{{x}+{C}}}$

And, move the 16 to the right side.

$\displaystyle{y}={{e}}^{{{x}+{C}}}-{16}$

Therefore, the general solution is $\displaystyle{y}={{e}}^{{{x}+{C}}}-{16}$ .

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