**(Update 3/18/17): **I wrote this over one year ago. These are only separable differential equation (all you have to do is put the y's on one side and put the x's on the other). I am currently writing more advanced differential equations posts here. Also, my posts look nice now (I didn't use LaTeX before).

# Solve a Differential Equation

## y' = 2x/y

yy' = 2x

Now, you can integrate both sides.

∫yy'dx = ∫2xdx

On the left side, let u = y, du = y'dx

∫udu = x^{2} + C

u^{2}/2 = x^{2} + C

y^{2}/2 = x^{2} + C

## y^{2} - 2x^{2} = C

# Growth and Decay Models

When something grows or decays exponentially, you can say y' = ky, so as y increases, y' increases (or decreases, depending on the k value). Now, let's solve it:

y' = ky

y'/y = k

Integrate both sides: ln|y| = kx + C

|y| = e^{c}e^{kx}

Note that the expression on the right always has to be positive. So, if you take out the absolute value, the expression can be negative. So, we can replace e^{c} with c (note: it's a different c, but it still serves as a constant)

## y = C*e^{kt}

## Example 1: Radioactive Decay

Pu-239's half life is 24,100 years. Now, suppose 10 grams of Pu-239 was released. How long will it take for the 10 grams to decay to 1 gram?

Now, we know it's proportional, so we could say y' = ky, or we could go right away to:

y = Ce^{kt} where t = time in years.

When t = 0, y = 10, 10 = C(1),**C = 10**

y = 10e^{kt}

Now, we know the half life is 24,100 years, so when t = 24100, y = 5

5 = 10e^{24100k}

1/2 = e^{24100k}

ln(1/2)/24100 = k**k = -0.0000288**

**y = 10e ^{-0.0000288t}**

1 = 10e^{-0.0000288t}

ln(1/10)/ln(1/2) * 24100 = t

t = 80,058 years

## Example 2: Newton's Law of Cooling

Newton's Law of Cooling basically says that a substance cools faster the further away it is from the surrounding temperature. For example, if a cup of tea is 120 degrees Fahrenheit, it will cool really quickly before slowing as it got closer to room temperature.

Question: Let y represent the temperature (F) of an object in a room whose temperature is kept at a constant 60 F. If the object cools from 100 F to 90 F in 10 minutes, how much longer will it take to cool to 80 F?

### Differential Equation

y' = k(y - 60) (Slows as it gets closer to 60 F)

y'/(y - 60) = k

ln|y-60| = kt + C

## y = 60 + Ce^{kt} (t is in minutes)

When t = 0, y = 100, so 100 = 60 + C, so C = 40.

When t = 10, y = 90, so 90 = 60 + 40e^{10k}

3/4 = e^{10k}

ln(3/4)/10 = k

y = 60 + 40e^{tln(3/4)/10}

y = 60 + 40(3/4)^{t/10}

1/2 = (3/4)^{t/10}

1/2 = 0.9716^{t}

t = 24.09 minutes

Answer = 24.09 - 10 =