Edexcel Further Pure 2

Transcription

1 Edecel Further Pure Second order differential equations Section : Non-homogeneous second order differential equations Notes and Eamples These notes contain subsections on Finding particular integrals Finding general and particular solutions Using substitutions Finding particular integrals The table below illustrates all the forms of particular integrals that you could meet. Function f() Particular integral Constant term c Linear function a + b Quadratic function a² + b + c Eponential function involving e p k e p Function involving sin p and / or cos p acos p bsin p The worked eamples in the tetbook illustrate all of these types of particular integral. Here is a further eample, illustrating an important point about the form of the particular integral. Eample Find a particular integral of the differential equation d² y y ² ² Since the function on the right-hand side is quadratic, use the trial function y a² b c a b Notice that even though the function on the right-hand side d² y a d² involves only a term in ², the trial function must also involve a term in and a constant term, as these will appear Substituting these into the differential equation: when differentiating. a ( a b) ( a² b c) ² MEI, 4/0/ /5

2 Edecel FP nd order DEs Sec. Notes & Eamples a² ( a b) (a b c) ² Equating coefficients of ²: a a Equating coefficients of : a b 0 b a 3 Equating constant terms: a b c 0 c ( a b) 3 The particular integral is y ² 4 4 A similar situation may arise when the particular integral involves trig functions. If the right-hand side of the differential equation is cos, but the particular integral must also involve sin, as this will appear when differentiating. If the form which you would epect the particular integral to take is the same as part of the complementary function, this is a problem, since this means that any particular integral satisfies the homogeneous equation and therefore does not satisfy the whole equation. For eample, you would epect the differential equation 4 4 e to have a particular integral of the form a e, but this will not work since it satisfies the equation 4 4 0,. In such cases, a particular integral of the form a e is usually used. However, in this case, this form is also the same as part of the complementary function (since the auiliary equation has a repeated root) so instead a particular integral of the form a e is used. This situation can also arise with trig functions, as shown in the eample below. Eample Find the general solution of the differential equation 4y cos sin. The auiliary equation is m 4 0 m i Complementary function is y Acos B sin The particular integral is of the form y ( acos bsin ) Since the general solution involves sin and cos, this form of the particular integral is used. MEI, 4/0/ /5

3 Edecel FP nd order DEs Sec. Notes & Eamples acos bsin ( asin b cos ) a sin b cos ( a sin b cos ) ( 4a cos 4b sin ) 4a sin 4b cos 4 ( a cos b sin ) Substituting into the differential equation: 4asin 4b cos 4 ( acos bsin ) 4 ( acos bsin ) cos sin 4asin 4b cos cos sin Equating coefficients of sin : 4a a Equating coefficients of cos : 4b b Particular integral is y 4sin 4cos General solution is y Acos Bsin sin cos Finding general and particular solutions You can see from the worked eamples in the tetbook that solving a differential equation of the form a b cy f ( ) is quite a lengthy process. However, although it is easy to make careless errors, the steps involved are all quite straightforward. Here is a summary of the process. Write down and solve the quadratic auiliary equation. Use the solutions of the auiliary equation to write down the complementary function in the appropriate form, with unknown constants A and B. Look at the function f() on the right-hand side of the differential equation. Check whether it is of the same form as any part of the complementary function, and then write down the appropriate form of the particular integral, which will involve one, two or three unknown constants. Differentiate the particular integral twice, and substitute into the original differential equation. Then equate coefficients to find the values of the unknown constants in the particular integral. Write down the general solution of the differential equation by adding the particular integral to the complementary function. If a particular solution is required, substitute the given conditions to find the values of the unknown constants A and B. (This may involve solving a pair of simultaneous equations). Here is a further eample. MEI, 4/0/ 3/5

4 Edecel FP nd order DEs Sec. Notes & Eamples Eample 3 Find the particular solution of the differential equation d² y y sin ² in the case where y = 0 and = 0 when = 0. The auiliary equation is: m m 0 ( m )² 0 m The complementary function is: y ( A B)e The particular integral is y asin bcos a cos bsin d² y asin bcos d² Substituting into the differential equation: asin bcos ( acos bsin ) asin bcos sin ( a b a)sin ( b a b)cos sin b sin acos sin Equating coefficients of cos : a 0 Equating coefficients of sin : b The particular integral is y cos The general solution is: y ( A B)e cos When = 0, y = 0 0 A A Be ( A B)e sin When = 0, = 0 0 B A B The particular solution is y ( )e cos For practice in finding particular solutions, try the interactive questions Second order differential equations (particular solutions). These involve only homogeneous equations for which the auiliary equation has two distinct roots. MEI, 4/0/ 4/5

5 Edecel FP nd order DEs Sec. Notes & Eamples You can eplore second order differential equations (homogeneous or nonhomogeneous) and their particular solutions using the Autograph resource Second order differential equations. Using substitutions The use of substitutions allows you to solve a wider range of differential equations, by changing a differential equation which you cannot solve into one which can be solved by one of the methods you have alrea met. Not all differential equations can be solved in this way, and it is not always obvious what substitution might work. However, you will only be epected to deal with situations in which you are given the substitution. MEI, 4/0/ 5/5