MAP 2302 Differential Equations Summary 9:Higher Order Differential Equations Sanchez DE equations reducible in order I. Case I. Equations of the form y (n) F( x, y (n 1) ) can be reduced to first order by the transforma tion w y (n 1) II . Case II . A second order differenti al equations of the form y F( y , y ) dw can be reduced to first order by the transforma tion w y and y w dy dy dy dw dw dy dw dw Note : w y y y w dx dx dx dy dx dy dy III. Reduction of order- D’Alembert’s Procedure. If y1 is a solution of the related homogeneous equation of a given differential equation, Then the substitution y=vy1 and then v’=w produces a linear differential equation of order n-1. Proof: (for n=2) 1. Let y Py Qy R be a differenti al equation 2. Let y Py Qy 0 be the related hom ogeneous differenti al equation 3. y1 is a solution of y Py Qy 0 y1 Py1 Qy 1 0 4. y vy1 y v y1 vy1 and y v y1 v y1 v y1 vy1 y v y1 2 v y1 vy1 5. y Py Qy R v y1 2 v y1 vy1 Pv y1 vy1 Qvy 1 R v y1 v ( 2y1 Py 1 ) vy1 Py1 Qy 1 R v y1 v ( 2y1 Py 1 ) v0 R Let w v and w v y1w ( 2y1 Py 1 )w R IV. Abel’s Formula: If y1 is a solution of a second order hom ogeneous linear differenti al equation , y P( x )y Q( x )y 0, then y 2 y1 p( x )dx e y1 2 -1- dx Linearly independent solutions of a differential equation Wronskian of a set of functions. The Wronskian of the set of functions f1(x), f2(x), f2(x),…, fn(x) is defined by the determinant f1 ( x) f 2 ( x) ... f n ( x) f1 ( x) f 2 ( x) ... f n ( x . . . . W (f1 , f 2 , . . ., f n ) provided the functions . . . . . . . . (n 1) f1 (n 1) ( x) f 2 ( x) (n 1) . . . f1 ( x) have n 1 derivative s. Test for independence of a set of functions: The set of functions f1 , f 2 , . . ., f n is linearly independen t if the Wronskian W(f1 , f 2 , . . ., f n ) 0 for every x in the int erval . Homogeneous Differential Equations of order n Linear Differential Operator of order n. If Dk y dk y k , then P an Dn an 1Dn 1 an 2 Dn 2 . . . a 2 D2 a1D a 0 dx is called a Linear Differential Operator of order n. Properties of the solutions of an nth order homogeneous linear differential equation: 1) Any linear combination of solutions is also a solution. That is, if y1 and y 2 are solutions and c1 anf c 2 are any cons tan ts, then c1y1 c 2 y 2 is also a solution 2) An nth order homogeneous linear differential equation has n linearly independent solutions: y1 , y 2 , y 3 , . . , y n . 3). The general solution of the homogeneous linear differential equation is given by y h c1y1 c 2 y 2 . . . c n y n . 4) If Yh is the homogeneous solution of a non-homogeneous differential equation and Yp is a particular solution of the non-homogeneous differential equation, then Y = Yh + Yp is the general solution of the non-homogeneous differential equation. -2- 5) Theorem. If y e ax is a solution of the hom ogeneous differenti al equation with constant coefficients an dn y dx n a n 1 d n 1y dx n 1 ... a2 d2y dx 2 a1 dy a o y 0, then " a" is a solution of dx the polynomial function a n x n a n 1x n 1 . . . a 2 x 2 a1x a 0 0 Definition : the polynomial function a n x n a n 1x n 1 . . . a 2 x 2 a1x a 0 0 is called the Characteri stic Equation of the linear hom ogeneous differenti al equation with constant coefficients. The solutions are called the Eigenvalue s of the DE. Definition : an Dn an 1Dn 1 . . . a 2 D2 a1D a 0 0 is called the differenti al operator of the linear hom ogeneous differenti al equation with constant coefficients . Theorem: If y e ax is a solution of the hom ogeneous differenti al equation with constant coefficients an dn y dx n a n 1 d n 1y dx n 1 ... a2 d2y dx 2 a1 dy a o y 0, then " a" is a solution of dx the polynomial function a n x n a n 1x n 1 . . . a 2 x 2 a1x a 0 0 Definition : the polynomial function a n x n a n 1x n 1 . . . a 2 x 2 a1x a 0 0 is called the Characteri stic Equation of the linear hom ogeneous differenti al equation with constant coefficients . The solutions are called the Eigenvalue s of the DE. Definition : a n D n a n 1D n 1 . . . a 2 D 2 a1D a 0 0 is called the differenti al operator of the linear hom ogeneous differenti al equation with constant coefficients . I. Theorem- If m is a double root of the characteristic polynomial of a Homogeneous Linear Differential Equation with constant coefficients then y1 e mx and y 2 xemx are linearly independen t solutions of the differenti al equation . Theorem . If m is a multiple root (k times ) of the characteri stic equation of a homogeneous linear differential equation with constant coefficients, then y (c1 c 2 x c3x 2 . . . ck xk 1 )emx is a hom ogeneous solution of the DE. -3- Theorem : If m1 i and m 2 i are conjugate complex roots of the characteri stic polynomial of a linear hom ogeneous DE with cons tan t coefficients, then y e x (c1 cos x c 2 sin x) is a solution of the DE. Definition: Boundary Value Problems (BVP): a linear differential equation of order two or greater in which the dependent variable y or its derivatives are specified at different points. A boundary value problem can have many, one, or no solutions. Properties of linear differential operators. 1. If P and Q are linear differential operators then PQy=P(Qy) 2. If P and Q are linear differential operators with constant coefficients then PQ=QP 3. Dm Dn Dm n 4. Dk e ax a k e ax 5. f ( D)e ax e ax f (a) 6a. First Shifting Formula D e ax y e ax D a y Note: 6b. e ax Dy ( D a) e ax y 7a. Second Shifting Formula D n e ax y e ax D a n y 7b. Note : e ax Dn y ( D a)n e ax y 8a. f (D) e ax y e ax f ( D a)y 8) e ax f ( D)y f ( D a) e ax y 9. f D a e ax y e ax f D y -4- Undeterminate Coefficients II. Un-determinate coefficients. This method can be applied if 1) The coefficients of the DE are constants 2) g(x) is a constant, a polynomial , an exponential function, a sine or a cosine function, or a finite sum or product of these functions. Solution: 1. Find the homogeneous solution Yh 2. The form of the particular solution Yp is a linear combination of all linearly independent functions that are generated by repeated differentiations of g(x) 3. If any Ypi contains terms that duplicate terms in the homogeneou s solution Yh , then Ypi must be multiplied by x n , where n is the smallest positive integer th at eliminates that duplicatio n. 4. Substitute the Yp obtained in the differential equation 5. Use the method of un-determinate coefficients to solve for the constants. 6. The non-homogeneous solution of the differential equation is y=Yh+Yp. Annihilators Annihilators. An annihilator of a function y=f(t) is a linear differential P(D) that satisfies the condition P(D)[f(t)] =0. This is the same as saying that f(t) is a solution of the homogenous differential equation P(D)f(t)=0. Finding the general form of the particular solution of a non-homogeneous linear differential equation by using annihilators. Step 1. Express the DE in linear differential form, that is, L(y)=g(x) Step 2. Find the homogeneous solution (complementary solution) of the differential equation, that is find the general solution of L(y)=0 Step 3.. Find an annihilator L1for g(x), that is L1(g(x)=0 Step 4. Operate on both sides of the non-homogeneous equation with the annihilator L1, that is, L1 L(y ) L1 (g(x) 0 -5- Step 5. Find the homogeneou s solution (complemen tary solution) of the differenti al equation L1 L( y ) 0 Step 6. The general for the particular solution of L( y ) g( x ) is given by e lim inating from the solution of L1 L( y ) 0, the terms which belong to the solution of L( y ) 0 Variation of parameters technique. Technique: Step 1. Let y h c1y1 c 2 y 2 be the hom ogeneous solution Step 2. Assume y p v1y1 v 2 y 2 where v1 and v 2 are functions and v1 y1 v 2 y 2 0 Step 3. Solve the system for v1 , v 2 v1 y1 v 2 y 2 0 v1 y1 v 2 y 2 g( x ) Step 4. Integrate each of v1 , v 2 , n to find v1 , v 2 Step 5. Find y p v1y1 v 2 y 2 Step 6. Find y y h y p Technique: The above technique can be generalized to a differential of order n. Step 1. Let y h c1 y1 c2 y 2 ... cn y n be the hom ogeneous solution Step 2. Assume y p v1 y1 v2 y 2 ... vn y n where vi are functions Step 3. Solve the system for v1, v2 , ..., vn v1 y1 v2 y 2 ... vn y n 0 v1 y1 v2 y 2 ... vn y n 0 v1 y1 v2 y 2 ... vn y n 0 v y ( n 1)1 v y ( n 1) 2 ... v y ( n 1) n g ( x) 2 n 1 an Step 4. Integrate each of v1, v2 , ..., vn to find v1 , v2 , ..., vn Step 5. Find y p v1 y1 v2 y 2 ... vn y n Step 6. Find y y h y p -6- Cauchy-Euler Differential Equation : Definition:A DE of the form an x n d (n ) y a n1x dx n n 1 d (n1) y 2 d2y dy . . . a2x a1x a 0 y g( x ) dx dx n1 dx 2 is called a Cauchy-Euler Differential Equation Theorem: If x 0, the substituti on x e t , t ln x, transforms the equation to a linear DE of order n with constant coefficients . If x 0, the substituti on x e t , t ln( x) transforms the equation to a linear DE of order n with constant coefficients . Technique: m Step 1. Write the characteri stic equation a k m (m 1)(m 2) . . . m k 1 0 k 0 and solve for m. Step 2. Follow the procedures used for Linear equations of order n with constant coefficients. Step 3. Write the answer in the form y=h(t). Step 4. If y f(x), let x e t and t ln x -7-