LINEAR DIFFERENTIAL EQUATIONS OF ORDER 1. where a(x) and b(x) are functions. Observe that this class of equations includes equations of the form

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1 LINEAR DIFFERENTIAL EQUATIONS OF ORDER 1 We consier ifferential equations of the form y + a()y = b(), (1) y( 0 ) = y 0, where a() an b() are functions. Observe that this class of equations inclues equations of the form a()y + b()y = c(), y( 0 ) = y 0, with a() 0 because we can ivie through by a() an obtain y + b() y = c(), a() a() y( 0 ) = y 0, We now eplain a metho to solve (1) in a systematic way calle the metho of Integrating Factors Let us consier the following eample Integrating factors Eample: Solve the following initial value problem (2) y + y =, y(0) = 6, In orer to solve (2) we follow the steps below Step 1 Multiply equation (2) through by e m() where m() is a function to be etermine, in other wors consier the equation (3) e m() y + e m() y = e m(). Step 2 Solve for m() so that we have e m() y + e m() y = ( (4) e m() y ). In orer to fin such function m(), we write out the right-han sie of (5) e m() y + e m() y = ( (5) e m() y ) = m ()e m() y + e m() y which simplifies to (6) e m() y = m ()e m() y, 1

2 2 LINEAR DIFFERENTIAL EQUATIONS OF ORDER 1 (7) y = m ()y, an this says that it suffices to choose m() to be any function satisfying m () =, for eample (8) m() = 2 2. Step 3 Insert the function m() given by (8) into (3) an use (5) to obtain ) (9) (e 2 2 y = e 2 2. Step 4 Integrate (9) an make use of the initial conition in (2). This means writing ) (e s2 2 y(s) s = se s2 2 s, s from where we conclue that 0 e 2 2 y y(0) = e }} y(0)= 6 y() = 7e Note that y() satisfies the initial conition of the problem since y(0) = = 6 an also so y() is inee a solution of (2). y + y = 7e 2 2 7e =, 0 The above solution seems to be lengthy but this is only because we have broken up the metho into many steps. For the net eample we give a more concise presentation of the metho. Eample: Solve the initial value problem (10) (11) which can also be rewritten as y = y + 2 sin() y() = 0. (12) y y = sin() (13) y() = 0.

3 LINEAR DIFFERENTIAL EQUATIONS OF ORDER 1 3 Let us start by multiplying equation (12) through by e m() an let us solve for m() so that e m() y em() y = ( e m() y ) = m e m() y + e m() y. Note that it suffices to choose m() so that m () = 1, for eample Let us rewrite (12) as (14) m() = ln(). ( ) e m()y = e m() sin(), ( ) 1 y() = sin(). We now integrate (14) using the initial conitions in (12), which gives us ( ) y(s) s = sin(s)s = (1 + cos()) s s We conclue that y() In general, for an equation of the form y() = (1 + cos()). } } y()=0 y() = (1 + cos()) y + a()y = b(), we use as integrating factor a function e m() with m () = a(). Metho of variation of parameters This is an alternative metho for solving linear ifferential equations of orer 1. Let us consier again the equation (15) y y = sin() (16) y() = 0.

4 4 LINEAR DIFFERENTIAL EQUATIONS OF ORDER 1 For a problem like (15) there are two equations that we can consier, namely the homogeneous equation an the inhomogeneous equation. The homogeneous equation is given by (17) y y = 0. an the inhomogeneous equation is given by y y = sin(). We highlight the above istinction with the table below: Homogeneous Equation Inhomogeneous Equation y y = 0 y y = sin() The metho of variation of parameters can be applie using the following steps Step 1 Fin the general solution of the homogeneous problem. By general solution we mean the solution obtaine without prescribing initial conitions. Let us solve the homogeneous problem by introucing logarithmic erivatives where M = e C is a constant. y = y, y y = 1 ln(y) = 1, ln(y) = ln() + C y = }} e C M y = M, Step 2 Consier a solution of the inhomogeneous equation of the form (18) y() = M()

5 (19) LINEAR DIFFERENTIAL EQUATIONS OF ORDER 1 5 where M() is a function of. If we insert this guess for y() into the inhomogeneous equation (15) we obtain Step 3 Integrate (19) an obtain so that sin() = y y = M() (M() ) = M () + M() M() From (18), our solution is M (s)s = = M (), M () = sin(). sin(s)s = (1 + cos()). M() = M() (1 + cos()). y() = M() = (M() (1 + cos())), an using the initial conition y() = 0 we have an M() = 0. It follows that which we know solves (15). 0 = y() = (M() (1 + cos())) = M() y() = (1 + cos()) Remark: The metho is calle variation of parameters because the general solution of the homogeneous problem (17) epens on a constant or parameter M an in orer to obtain a solution of the inhomogeneous problem (15) we replace the parameter M by a function of to be etermine. In other wors, in orer to solve the inhomogeneous problem we let the parameter M vary.