Page 25

110

CHAPTER 2

24.

Linearity and Nonlinearity

SECTION 2.2

Solving the First-Order Linear Differential Equation

y ′ − y = e3t Here p ( t ) = −1, therefore the integrating factor is μ ( t ) = e ∫

p ( t ) dt

− dt = e ∫ = e −t .

Multiplying each side of the equation y ′ − y = e3t by e −t yields

(

)

d ye−t = e 2t . dt 1 1 Integrating gives ye−t = e 2t + c . Solving for y gives y ( t ) = cet + e3t . 2 2 25.

y ′ + y = sin t Here p ( t ) = 1 therefore the integrating factor is μ ( t ) = e ∫

p ( t ) dt

= e ∫ = et . dt

Multiplying each side of the equation y′ + y = sin t by et gives

( )

d yet = et sin t . dt 1 1 Integrating gives yet = et ( sin t − cos t ) + c . Solving for y gives y ( t ) = ( sin t − cos t ) + ce −t . 2 2 26.

y′ + y =

1 1 + et

Here p ( t ) = 1 therefore the integrating factor is μ ( t ) = e ∫ Multiplying each side of the equation y′ + y =

p ( t ) dt

= e ∫ = et . dt

1 t yields t by e 1+ e

( )

d et ye t = . dt 1 + et

(

)

(

)

2tdt

= et .

Integrating gives yet = ln 1 + et + c . Solving for y gives y ( t ) = e −t ln 1 + et + ce −t . 27.

y′ + 2ty = t Here p ( t ) = 2t , therefore the integrating factor is μ ( t ) = e ∫

p ( t ) dt

Multiplying each side of the equation y ′ + 2ty = t by et yields 2

= e∫

2

110

Solutions manual for differential equations and linear algebra 2nd edition by farlow ibsn 9780134689  

Solutions Manual for Differential Equations and Linear Algebra 2nd Edition by Farlow IBSN 9780134689548 Full clear download( no error format...

Solutions manual for differential equations and linear algebra 2nd edition by farlow ibsn 9780134689  

Solutions Manual for Differential Equations and Linear Algebra 2nd Edition by Farlow IBSN 9780134689548 Full clear download( no error format...

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