1. Differential Equations

2. We know that a differential equation (DE) connects a rate of change to another variable Slope = dy/dx Velocity = ds/dt Acceleration = dv/dt If we have the rate of change how do we get back to the original quantity?
Integrate

3. Since we can’t find c we say that this is a general solution to the DE
Eg 1 V = 8t3 -3t Find the distance equation distance = ∫v(t)dt = ∫ (8t3 -3t2 + 4) dt = 2t4 – t3 + 4t + c
Since we can’t find c we say that this is a general solution to the DE
If we had further info and could find c then this would give a particular solution to the DE

4. So the solution to a DE is found by integrating and is a function (not a number)
Function = y
integrate
differentiate
DE = Rate of change = dy/dx

5. This is a 1st order DE since we have the1st derivative
We need to be able to: solve DE’s (look at later) verify that a given function is a solution to a DE
To verify that a given function is a solution to a DE we differentiate eg. Verify that y = ½ x2 is a solution to the DE dy/dx – x = 0
This is a 1st order DE since we have the1st derivative

6. This is a 2nd order DE since we have the 2nd derivative
EG 2 Verify that y = excosx satisfies the DE
This is a 2nd order DE since we have the 2nd derivative