1 / 18

# Reminder siny = e x + C This must be differentiated using implicit differentiation

Implicit Differentiation. Reminder siny = e x + C This must be differentiated using implicit differentiation When differentiating y’s write dy When differentiating x’s write dx Divide by dx. Divide by dx. Rearrange to make the subject. Implicit Differentiation.

Télécharger la présentation

## Reminder siny = e x + C This must be differentiated using implicit differentiation

E N D

### Presentation Transcript

1. Implicit Differentiation Reminder siny = ex + C This must be differentiated using implicit differentiation • When differentiating y’s write dy • When differentiating x’s write dx • Divide by dx

2. Divide by dx Rearrange to make the subject Implicit Differentiation For example siny = ex + C cosy dy = ex dx

3. multiply by dx multiply by cosy integrate Reversing the Process Integrating Differentiating siny = ex + C cosy dy = ex dx cosy dy = ex dx siny = ex + C

4. Finding the constant C siny = ex + CTo find the constant C a boundary condition is needed. If we are told that when x = 0 then y = p/2then we can find C. siny = ex + C Substitute x = 0 and y = p/2 sin p/2 = e0 + C So C = 0 siny = ex y = sin-1(ex)

5. 1st Order Differential Eqns Separating the Variables (C4) 1) Collect the x’s on one side and the y’s and the on the other side. 2) Multiply both side by dx and integrate. 3) Simplify the resulting equation. y fns join the dy and x fns join the dx usually by Cross Multiplying N.B y functions are integrated w.r.t y x functions are integrated w.r.t x

6. Ex1 Solve the differential equation Multiply both sides by y Multiply both side by dx and integrate Find C by substituting in 2 boundary conditions

7. Ex1 Differential equations where the variables cannot be separated If we try to rearrange so that the x`s and dx`s are on one side and the y`s and dy`s are on the other it is impossible. Try it!!

8. This is an example of a 1st order Linear Differential Equation. It can be solved however by multiplying every thing by x3.

9. What could we differentiate that would give this expression. It must contain a mixture of x`s and y`s…….. Differentiating x3y x3dy + 3x2ydx using implicit differentiation divide by dx

10. So if we integrate we obtain x3y Integrate both sides

11. If Then the integrating factor = How do we find what to multiply by to make the LHS into an expression which can be integrated by inspection. THE INTEGRATING FACTOR !!!!!!

12. Differential equations where the variables cannot be separated Steps 1. Rearrange in the form 2. Find the integrating factor I.F. = 3. Multiply every term by the I.F. 4. Integrate each side – the integral of the L.H.S. is I.F × y

13. 1. Rearrange in the form Divide by x P(x) = and Q(x) = 1

14. P(x) = I.F = 3. Multiply every term by the I.F. 4. Integrate each side – the integral of the L.H.S. is I.F × y 2. Find the integrating factorI.F. =

15. If we integrate How can you spot what the integral of the LHS is??? we obtain x3y This is the same as I.F. × y So the integral of the LHS is I.F. × y

16. + cotx y  = cosecx x = 0 when y = 0 + sinx cotx y  = sinx cosecx 1. Rearrange in the form P(x) = cotx Q(x) = cosecx = sinx I.F = 2. Find the integrating factorI.F. = 3. Multiply every term by the I.F.

17. + sinx y  = sinx + cosx y  = 1 + sinx cotx y  = sinx cosecx sinx y = Simplify 4. Integrate each side – the integral of the L.H.S. is I.F × y Integrating the LHS gives sinx y i.e I.F × y = x + c

18. sinx y = = x + c Substitute x = 0 and y = 0 gives c = 0 sinx y = x So y =

More Related