# What is a solution to the differential equation dP-6Pdt=0 with P=5 when t=0?

Jun 27, 2016

$P = 5 {e}^{6 t}$

#### Explanation:

$\mathrm{dP} - 6 P \mathrm{dt} = 0$

so separate it!!

$\mathrm{dP} = 6 P \mathrm{dt}$

$\frac{\mathrm{dP}}{P} = 6 \mathrm{dt}$

$\int \frac{\mathrm{dP}}{P} = 6 \int \mathrm{dt}$

$\ln P = 6 t + \alpha$

$P = {e}^{6 t + \alpha} = {e}^{6 t} {e}^{\alpha} = \beta {e}^{6 t}$

NB: we have just replaced the constant ${e}^{\alpha}$ by $\beta$, a simpler form of constant, so $\beta = {e}^{\alpha}$

with final form $P = \beta {e}^{6 t}$, we can plug in the initial value so

$5 = \beta {e}^{6 \times 0} = \beta$

so we have

$P = 5 {e}^{6 t}$