# Intro - What is a diﬀerential equation A diﬀerential...

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Unformatted text preview: What is a diﬀerential equation? A diﬀerential equation is any equation containing one or more derivatives. Notations of derivatives: ♠ Ordinary derivatives: let y be a function of x, the derivatives of y are denoted by dy ; dx d2 y (2) second derivative: y or 2 ; dx dn y (3) n-th derivative: y (n) or n . dx (1) ﬁrst derivative: y or ♠ Partial derivatives: let u be a function of x and y , the derivatives of u are denoted by ∂u ∂u , uy or ; ∂x ∂y ∂ 2u ∂ 2u ∂ 2u , uyy or , uxy or . (2) second derivative: uxx or ∂x2 ∂y 2 ∂x∂y (1) ﬁrst derivative: ux or Examples of diﬀerential equations: (1) y = x; (2) y + xy + y = 0; (3) uxx + uyy = 0. 1 &ODVVLILFDWLRQ RI 'LIIHUHQWLDO (TXDWLRQV 2UGLQDU\ YV SDUWLDO GLIIHUHQWLDO HTXDWLRQV \$Q RUGLQDU\ GLIIHUHQWLDO HTXDWLRQ 2'( LV D GLIIHUHQWLDO HTXDWLRQ ZLWK D VLQJOH LQGHSHQGHQW YDULDEOH VR WKH GHULYDWLYH V LW FRQWDLQV DUH DOO RUGLQDU\ GHULYDWLYHV \$ SDUWLDO GLIIHUHQWLDO HTXDWLRQ 3'( LV D GLIIHUHQWLDO HTXDWLRQ ZLWK WZR RU PRUH LQGHSHQGHQW YDULDEOHV VR WKH GHULYDWLYH V LW FRQWDLQV DUH SDUWLDO GHULYDWLYHV 2UGHU RI D GLIIHUHQWLDO HTXDWLRQ 7KH RUGHU RI D GLIIHUHQWLDO HTXDWLRQ LV HTXDO WR WKH RUGHU RI WKH KLJKHVW GHULYDWLYH LW FRQWDLQV ([DPSOHV  \  \ W  Hí W  FRV W \ í VLQ W \  \ í \  \  \   \   X[[  \  í \ \  \ ILUVW RUGHU 2'( W FRV W HW FRV W  ILUVW RUGHU 2'( VHFRQG RUGHU 2'( IRXUWK RUGHU 2'( XWW  XW VHFRQG RUGHU 3'(  H W WKLUG RUGHU 2'( /LQHDU YV QRQOLQHDU GLIIHUHQWLDO HTXDWLRQV \$Q QWK RUGHU RUGLQDU\ GLIIHUHQWLDO HTXDWLRQ LV FDOOHG OLQHDU LI LW FDQ EH ZULWWHQ LQ WKH IRUP \ Q DQí W \ Qí  DQí W \ Qí  «  D W \  D W \  J W  :KHUH WKH IXQFWLRQV D¶V DQG J DUH DQ\ IXQFWLRQV RI WKH LQGHSHQGHQW YDULDEOH W LQ WKLV LQVWDQFH 1RWH WKDW WKH LQGHSHQGHQW YDULDEOH FRXOG DSSHDU LQ DQ\ VKDSH RU IRUP LQ WKH HTXDWLRQ EXW WKH GHSHQGHQW YDULDEOH \ DQG LWV GHULYDWLYHV FDQ RQO\ DSSHDU DORQH LQ WKH ILUVW SRZHU QRW LQ D GHQRPLQDWRU RU LQVLGH DQRWKHU WUDQVFHQGHQWDO IXQFWLRQ ,Q RWKHU ZRUGV WKH ULJKWKDQG VLGH RI WKH HTXDWLRQ DERYH PXVW EH D OLQHDU IXQFWLRQ RI WKH GHSHQGHQW YDULDEOH \ DQG LWV GHULYDWLYHV 2WKHUZLVH WKH HTXDWLRQ VDLG WR EH QRQOLQHDU ,Q WKH H[DPSOHV DERYH  DQG  DUH OLQHDU HTXDWLRQV ZKLOH    DQG  DUH QRQOLQHDU  LV D OLQHDU SDUWLDO GLIIHUHQWLDO HTXDWLRQ DV HDFK RI WKH SDUWLDO GHULYDWLYHV DSSHDUV DORQH LQ WKH ILUVW SRZHU 7KH QH[W H[DPSOH ORRNV VLPLODU WR   EXW LW LV D VHFRQG RUGHU QRQOLQHDU HTXDWLRQ LQVWHDG :K\"  \ í \  \ HW FRV \ ([HUFLVHV  ±  'HWHUPLQH WKH RUGHU RI HDFK HTXDWLRQ EHORZ \$OVR GHWHUPLQH ZKHWKHU HDFK LV D OLQHDU RU QRQOLQHDU HTXDWLRQ  \  W \ FRV W  \  \ í \  HíW \  \ W\  \ W\  \  VHF W  \  \  \  \  í \   \ FRV \ WVLQ W \   HW \   \ í FRW HW \  )RU ZKDW YDOXH V RI Q ZLOO WKH IROORZLQJ HTXDWLRQ EH OLQHDU" \ í \ Q WQ VLQ QW \ OQ W \  \ í H W \ W  \ \$QVZHUV  VW RUGHU OLQHDU  UG RUGHU OLQHDU  QG RUGHU QRQOLQHDU  VW RUGHU OLQHDU  WK RUGHU QRQOLQHDU  QG RUGHU OLQHDU  VW RUGHU QRQOLQHDU  WK RUGHU QRQOLQHDU  WK RUGHU OLQHDU  :KHQ Q  RU  WKH HTXDWLRQ LV OLQHDU Some Formulae From Calculus ♠ Derivatives: (tn ) = ntn−1 ; (cos t) = − sin t; (sin t) = cos t; ( e t ) = et ; 1 (ln t) = (t > 0). t ♠ Antiderivatives/Indeﬁnite integrals: tn dt = tn+1 + C, (n = −1); n+1 cos t dt = sin t + C ; sin t dt = − cos t + C ; et dt = et + C ; 1 dt = ln |t| + C (t = 0). t ♠ Some properties of antiderivatives: [f (t) + g (t)] dt = af (t) dt = a Substitution : f (t) dt + g (t) dt; f (t) dt, where a is a constant; t=h(s) f (t) dt = = = == f (h(s))h (s) ds First Order Linear ODE A ﬁrst order linear ordinary diﬀerential equation can be written in the form y’=a(t)y+g(t), where a and g are arbitrary given functions of t. We can also write it as y’+p(t)y=g(t), where p(t) = −a(t). This is called the standard or canonical form of the ﬁrst order linear equation. We’ll start by attempting to solve a couple of very simple equations of such type: Case 1: a(t) = 0. In this event we get y = g ( t) , which is just a usual integration problem. Then the general solution of this equation is given by y= Comment: The antiderivative g (t) dt, g (t) dt actually represents a collec- tion of functions whose derivative is g (t). Each such function diﬀers from others by one (or more) constant. For example, t2 dt = t3 +C 3 where C is an arbitrary constant. Due to this fact, there are inﬁnitely many solutions for this equation. Examples: Solve the following 1st order linear ODE: (1) y = 10t4 + 3t2 ; 1 (2) y = e3t + , where t = 0; t (3) y = sin t + cos(2t). 3 Answers: (1) y = 2t5 + t3 + C ; 1 (2) y = e3t + ln |t| + C ; 3 1 (3) y = − cos t + sin(2t) + C . 2 Case 2: g (t) = 0. We have y = a(t)y. One trivial solution is given by y = 0. Assume y = 0. Rewrite the equation by dy = a(t)y. dt Move y to the left, and t to the right, we obtain dy = a(t)dt, y then integrate both sides, ln |y | = dy = y a(t) dt, and therefore we get |y | = e a(t) dt . () Comment: The above derivation is a simple example of Separation of Variables, whose basic idea is to separate dependent and independent variables to diﬀerent sides and then integrate. This method works for many types of equations and shall be used frequently in our course. We will see this technique in later sections. Now let us combine all these solutions in one formula. Pick an antiderivative of a(t), say A(t), then a(t) dt = A(t) + C , and |y | = eC eA(t) = C1 eA(t) , where C1 = eC is an arbitrary but positive number. In fact, we can drop the absolute value sign on the left, then C1 could be any positive 4 or negative (but nonzero) constant. So we rewrite ( ) as y = C1 e A ( t ) , C 1 = 0. When C1 = 0, the above formula indeed gives the trivial solution y = 0. To sum up, the general solution for y = a(t)y is given by y = CeA(t) , for any constant C, where A(t) is an antiderivative of a(t). Examples: Solve the following 1st order linear ODE: (1) y = ty ; 1 (2) y = y, t = 0; t (3) y = 2y − 4. Answers: 12 (1) y = Ce 2 t ; (2) y = C |t|, t = 0; (3) y = Ce2t + 2. (Do the substitution z = y − 2) Examples: Use the method of separation of variables to solve the following ODE: (1) y = 2y − 4; (2) y = t2 e−y . Answers: (1) y = Ce2t + 2; t3 +C . 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## This note was uploaded on 01/29/2012 for the course MATH 251 taught by Professor Chezhongyuan during the Spring '08 term at Penn State.

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