Differential Equations

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

Since we can’t find c we say that this is a general solution to the DE Eg 1 V = 8t3 -3t2 + 4. 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

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

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

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