Eact Equations An eact equation is a first order differential equation that can be written in the form M(, + N(,, provided that there eists a function ψ(, such that = M (, and N(, = Note : Often the equation is written in the alternate form of M(, d + N(, d Theorem (Verification of eactness): An equation of the form M(, + N(, is an eact equation if and onl if M = N Note : If M() is a function of onl, and N( is a function of onl, then M N triviall = Therefore, ever separable equation, M() + N(, can alwas be written, in its standard form, as an eact equation 008, 0 Zachar S Tseng A- -
The solution of an eact equation Suppose a function ψ(, eists such that (, = M and N (, = Let be an implicit function of as defined b the differential equation M(, + N(, () Then, b the Chain Rule of partial differentiation, d d ψ (, ( )) = + = M (, + N(, d d As a result, equation () becomes d ψ (, ( )) d Therefore, we could, in theor at least, find the (implicit) general solution b integrating both sides, with respect to, to obtain ψ(, = C Note : In practice ψ(, could onl be found after two partial integration steps: Integrate M (= ψ ) respect to, which would recover ever term of ψ that contains at least one ; and also integrate N (= ψ ) with respect to, which would recover ever term of ψ that contains at least one Together, we can then recover ever non-constant term of ψ Note 4: In the contet of multi-variable calculus, the solution of an eact equation gives a certain level curve of the function z = ψ(, 008, 0 Zachar S Tseng A- - 4
Eample: Solve the equation ( 4 ) + 4 First identif that M(, = 4, and N(, = 4 Then make sure that it is indeed an eact equation: M N = 4 and = 4 Finall find ψ(, using partial integrations First, we integrate M with respect to Then integrate N with respect to 4 4 M (, d= ( ) d= + ψ, = C ( ( 4 N (, d= 4 d= + ψ, = C ( ) ( Combining the result, we see that ψ(, must have non-constant terms: 4 and That is, the (implicit) general solution is: 4 = C, Now suppose there is the initial condition ( ) = To find the (implicit) particular solution, all we need to do is to substitute = and = into the general solution We then get C = 4 Therefore, the particular solution is 4 = 4 008, 0 Zachar S Tseng A- - 5
Eample: Solve the initial value problem ( cos( + + ) d+ ( cos( + ln + e ) d= 0, () First, we see that M (, = cos( + + and N (, = cos( + ln + e Verifing: M = sin( + cos( + = N = sin( + cos( + Integrate to find the general solution: ψ (, = cos( + + d= sin( + ln + + C(, as well, ( cos( ) ln e d sin( ) ln e = + + = + + + C ( ) ψ (, ) Hence, sin + ln + e + = C Appl the initial condition: = and : C = sin 0 + 0 ln () + e 0 + = The particular solution is then sin + ln + e + = 008, 0 Zachar S Tseng A- - 6
Eample: Write an eact equation that has general solution e + 4 4 6 = C We are given that the solution of the eact differential equation is ψ(, = e + 4 4 6 = C The required equation will be, then, simpl M(, + N(,, such that (, = M and N(, = Since = = e 4 e + 4 + 4 4, and 6 Therefore, the eact equation is: ( e + 4 4 ) + ( e + 4 4 6 ) 008, 0 Zachar S Tseng A- - 7
Summar: Eact Equations M(, + N(, Where there eists a function ψ(, such that = M (, and N(, = Verification of eactness: it is an eact equation if and onl if M = N The general solution is simpl ψ(, = C Where the function ψ(, can be found b combining the result of the two integrals (write down each distinct term onl once, even if it appears in both integrals): ψ (, = M (, d, and ψ (, = N(, d 008, 0 Zachar S Tseng A- - 8
Eercises A-: Write an eact equation that has the given solution Then verif that the equation ou have found is eact It has the general solution tan + 4 = C It has a particular solution ln + 5 = 9 0 For each equation below, verif its eactness then solve the equation + cos( ) + 4 4 4 4 + (4 + + 5) 5 ( + ( ), (0) = 5 6 ( + + 4 e ) + ( + e ), () 7 (5 e ) + ( 5 e ), (0) = 4 sin cos 8 ( + ) + ( ), (0) = 9 ( + ) + (arctan( ) ) 4, () = + 0 sin()sin( + cos() + (cos()cos( + sin()), (π/) = π Find the value(s) of λ such that the equation below is an eact equation Then solve the equation 5 6 (λ ) + ( λ) (λ sec ( λ ) + ( sec ( λ (0 4 6 + 6 sin( )) + (40 + λ cos( )) 008, 0 Zachar S Tseng A- - 9
Answers A-: ( tan + ) + ( sec 6 4 ) ( ) + ( + 5) + + sin( ) = C 4 4 4 + 5= C 5 + = 5 6 + + 4 e = 7 7 5 5 e = 4 cos 8 + = π 9 arctan( ) + = + 4 0 cos()sin( + sin() = π λ = ; 6 + = C λ = ; tan( = C λ ; 0 4 cos( ) = C 008, 0 Zachar S Tseng A- - 0