MAP- EXACT EQUATIONS AND INTEGRATING FACTORS First-order Differential Equations for Which We Can Find Eact Solutions Stu the patterns carefully. The first step of any solution is correct identification of the type of differential equation.. Total Differential of a Function F(, F is a function of two variables which has continuous partial derivatives over a domain D. The total differential of F is defined as : df(, F(, F(, + y (. D. Eact Differential M(, + N(, is an eact differential over D if F(, such that M (, + N(, df(, In other words, M(, N(, F(, F(, y. Eact Differential Equation M(, + N(, is an eact differential equation if M(, + N(, is an eact differential. Given a DE that can be written in the form M(, + N(,, find F(, such that df (, M(, + N(,.
MAP- The following theorem supplies a method for determining whether or not a DE is eact. Theorem. M(, and N(, are functions with continuous first partial derivatives in the domain D. M(, and N(, form the DE The above DE is eact in D if and only if M(, + N(, y M(, N(, (, D 4. Solution Method for Eact Equations a. Show M (, N (, y ) y F(, b. Set M(,. Integrate with respect to to get F (, F (, M (, + φ ( c. Differentiate with respect to y to get N(, [ M (, + φ ( ] N (, y ) y d. Solve for φ ( The answer will be of the form: F (, y ) g(, + φ (, if no boundary value is given.
MAP- Eercises: Which of these equations are eact? a. ( y + ) + ( y 4) b. ( y + ) ( + c. ( θ + )cos rdr + θ sin rdθ d. ( + ) + ( + y y e. / y + ( ) + ( y ) / Eercises: Solve y ( ) a. ( y y + ) + ( y y + ) given b. ( ye + e + y ) + ( e + c. Find A so that this equation is eact. ( A y + y ) + ( + 4
MAP- Integrating Factors M(, N(, What if y We may be able to rewrite the equation so that it is eact.. Definition If is not eact, but M (, + N(, μ (, M(, + μ(, N(, is eact, then μ (, is called an integrating factor. Eample Show that ( y ) + y - is not eact, then find n such that y n is an integrating factor. y a. ( ) y + y y + ( ) therefore the DE is not eact. b. Multiply the DE by y n, then solve. n n y ( y + y y n+ n+ n ( + ) ( + ) + ( + ) y y n y n y n must equal ( y n ) y n n y + n which means ( ) + must equal and n n ( n+ ) y must equal y c. Now, solve the equation. ( ) y y + y y ( ) Solution: for this to be so n must equal - ( ) ( ) + y + y > F (, + y + c 4
MAP-. Review: a. What is a linear DE? b. What is a first order DE? c. What is an eact DE of order? d. What is an integrating factor?. Eercise: a. Eamine 4 y + ( + ). Is it linear? order? eact? b. Multiply the given DE by y ( + ) the result. and answer the same questions about 4. Separable Equations F( ) G( + f ( ) g( ) This type of DE is called separable because it can be written in the form (variables can be separated) M ( ) + N( The first equation is usually not eact but multiplying it by the appropriate integrating factor will make it eact, but use of an integrating factor may eliminate solutions or may lead to etraneous solutions. After multiplying by the integrating factor F( ) g( + f( ) G( f ( G ) ( the equation becomes: where F( ) M( ) and f ( ) g( N( ). Gy ( ) where Solutions are of the form M ( ) + N( y ) + c f( ) & G(. 5
MAP- 5. Eercises: Which equations are separable? For each separable equation find the correct integrating factor. ( y + + y + ) + ( + ) a. Hint: factor y + + y + b. csc y+ sec v c. ( ) v e + cos u du + e (sin u + ) dv d. ( + e. vdu+ ( u uv) dv Homogeneous DE If M(, + N(, can be written in the form f (, where y n f (, g then the DE is homogeneous. If F(t, t t F(, then F is homogeneous of degree n. A homogeneous equation can be transformed to a separable equation by a change of variable y v. The new equation is separable in v and, so it can be solved. dv Given M (, + N(, is homogeneous, let y v. Then v +. Since the given DE is homogeneous, we know it can be written in the form y g. Since y v, y v g g g( v). 6
MAP- And since y g dv v+ g( v). Separating variables : + dv v g(v) v + dv g() v [ v g( v)] + dv : To solve, integrate: + dv c v ( g( ). Eercises: Solve y ln + F ln + Fv ( ) c or c a. ( + + ( ) b. y+ ( c. ( y+ ) + y( + 4) where y( ) 7
MAP- A Linear First Order DE with Variable Coefficients can be written in the following form. Py ( ) Q ( ) + This equation is eact only when P(). Proof: Write DE in the form M(, + N(, Py ( ) Q ( ) + + P( ) y Q( ) Py ( ) Q ( ) + [ ] But [ ] Py ( ) Q ( ) + M ( y, ) P ( ) Q ( ) P ( ) y y N(, () P() must equal for the DE to be eact. [ P ()y Q() ] The equation + can be solved though by finding a proper integrating factor. The factor depends only on, so call it μ (). 8
MAP- The new equation [ ( )P()y μ()q() ] + μ() μ is eact so ( ) Py ( ) ( Q ) ( ) ( P ) ( ) y μ μ μ Since [ ] d ( ) μ( ) [ μ ] d μ( )( ) μ( ) or dμ μ P() P( ) dμ μ P ( ) dμ μ P ( ) ln μ P( ) μ( ) e assuming μ ( ) > The solution of the DE is therefore of the form. Eercise: y e e Q( ) + c P( ) P( ) y + 4 9
MAP- Bernoulli Equations An equation of the form + P()y n Q()y is called a Bernoulli Equation. n Bernoulli equations can be transformed to linear equations by v y. Proof: Py ( ) Qy ( ) n + (y -n ) multiply n n y + P( ) y Q( ) Let v y n dv n dv ( ) n y P( v ) Q ( ) ( n) + y n dv n ( ) (-n) multiply ( ) dv n P ( ) v ( n) Q( + ) Let P( ) ( n) P( ) dv + P ( v ) Q( ) which is linear in v! Q ( ) ( n) Q( ) ( + ) + 4y y. Eample: change to Bernoulli 4 4 + y y where P ( ) + + + & Q ( ) + since y n y y - multiply y 4 + + +
MAP- Let n ( ) v y y y Solving for y v /, dv y Solving for dv y v dv 4 v + + + / / dv 4 + 4 dv v + v / dv ln v ln y v separate and integrate 4 + + + + ln y + + ln arctan Solving for y y e arctan ( + ) y Eercise: +