Laplace Transforms and Integral Equations Bernd Schr oder logo1 - PowerPoint PPT Presentation

Transforms and New Formulas An Example Double Check Laplace Transforms and Integral Equations Bernd Schr¨ oder logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain ( t ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain ( t ) Original DE & IVP logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain ( t ) L Original ✲ DE & IVP logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain ( t ) L Original Algebraic equation for ✲ DE & IVP the Laplace transform logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain ( t ) Transform domain ( s ) L Original Algebraic equation for ✲ DE & IVP the Laplace transform logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain ( t ) Transform domain ( s ) L Original Algebraic equation for ✲ DE & IVP the Laplace transform Algebraic solution, partial fractions ❄ logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain ( t ) Transform domain ( s ) L Original Algebraic equation for ✲ DE & IVP the Laplace transform Algebraic solution, partial fractions ❄ Laplace transform of the solution logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain ( t ) Transform domain ( s ) L Original Algebraic equation for ✲ DE & IVP the Laplace transform Algebraic solution, partial fractions ❄ L − 1 Laplace transform ✛ of the solution logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Everything Remains As It Was No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain ( t ) Transform domain ( s ) L Original Algebraic equation for ✲ DE & IVP the Laplace transform Algebraic solution, partial fractions ❄ L − 1 Laplace transform Solution ✛ of the solution logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check The Laplace Transform of an Integral logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check The Laplace Transform of an Integral � t 1. Definite integrals of the form 0 f ( τ ) d τ arise in circuit theory: The charge of a capacitor is the integral of the current over time. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check The Laplace Transform of an Integral � t 1. Definite integrals of the form 0 f ( τ ) d τ arise in circuit theory: The charge of a capacitor is the integral of the current over time. (We assume the capacitor is initially uncharged.) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check The Laplace Transform of an Integral � t 1. Definite integrals of the form 0 f ( τ ) d τ arise in circuit theory: The charge of a capacitor is the integral of the current over time. (We assume the capacitor is initially uncharged.) � � t � = F ( s ) 2. L 0 f ( τ ) d τ s logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Solve the Initial Value Problem � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = t , f ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Solve the Initial Value Problem � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = t , f ( 0 ) = 0 � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = f ( 0 ) = 0 t , logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Solve the Initial Value Problem � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = t , f ( 0 ) = 0 � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = f ( 0 ) = 0 t , sF + F − 2 F 1 = s 2 s logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Solve the Initial Value Problem � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = t , f ( 0 ) = 0 � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = f ( 0 ) = 0 t , sF + F − 2 F 1 = s 2 s � � s + 1 − 2 1 = F s 2 s logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Solve the Initial Value Problem � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = t , f ( 0 ) = 0 � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = f ( 0 ) = 0 t , sF + F − 2 F 1 = s 2 s � � s + 1 − 2 1 = F s 2 s Fs 2 + s − 2 1 = s 2 s logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check Solve the Initial Value Problem � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = t , f ( 0 ) = 0 � t f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = f ( 0 ) = 0 t , sF + F − 2 F 1 = s 2 s � � s + 1 − 2 1 = F s 2 s Fs 2 + s − 2 1 = s 2 s s = F s 2 ( s − 1 )( s + 2 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check � t Solve f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = t , f ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations

Transforms and New Formulas An Example Double Check � t Solve f ′ ( t )+ f ( t ) − 2 0 f ( z ) dz = t , f ( 0 ) = 0 s 2 ( s − 1 )( s + 2 ) = A s s + B C D = s 2 + s − 1 + F s + 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Laplace Transforms and Integral Equations