Differential Equations I – Final Exam Review

Linear Equations:

Form: 𝑎1(𝑥)𝑑𝑦

𝑑𝑥= 𝑎0(𝑥)𝑦 = 𝑓(𝑥)

Standard form : 𝑑𝑦

𝑑𝑥+ 𝑝(𝑥)𝑦 = 𝑞(𝑥) , where 𝑝(𝑥) =

𝑎0(𝑥)

𝑎1(𝑥) and 𝑞(𝑥) =

𝑓(𝑥)

𝑎1(𝑥)

If in standard form: multiply through by µ(𝑥) = 𝑒∫ 𝑝(𝑥)𝑑𝑥

Example: 𝑥𝑑𝑦

𝑑𝑥+ 2𝑦 = 𝑥−3

𝑑𝑦

𝑑𝑥+

2𝑦

𝑥= 𝑥−4 => µ(𝑥) = 𝑒∫

2

𝑥𝑑𝑥 = 𝑒2 ln|𝑥| = 𝑒ln|𝑥2| = 𝑥2

=> 𝑥2 𝑑𝑦

𝑑𝑥+ 2𝑥𝑦 = 𝑥−2 =>

𝑑

𝑑𝑥(𝑥2y) = 𝑥−2 => 𝑥2𝑦 = ∫ 𝑥−2𝑑𝑥 => 𝑥2𝑦 = −𝑥−1 + 𝐶

=> solution: 𝑦 = −𝑥−3 + 𝐶𝑥−2 = 𝐶

𝑥2 −1

𝑥3

Separable Equations:

Form: 𝑑𝑦

𝑑𝑥= 𝑓(𝑦)𝑔(𝑥)

Get all y dependence on one side and x dependence on the other, then integrate and solve for y.

Example: 𝑑𝑦

𝑑𝑥=

sec2 𝑦

1+𝑥2

𝑑𝑦

sec2 𝑦=

𝑑𝑥

1+𝑥2 => cos2 𝑦 𝑑𝑦 =𝑑𝑥

1+𝑥2 => ∫ cos2 𝑦 𝑑𝑦 = ∫1

1+𝑥2 𝑑𝑥 => ∫1+cos 2𝑦

2𝑑𝑦 = ∫

1

1+𝑥2 𝑑𝑥

=> 1

2𝑦 +

1

4sin(2𝑦) = tan−1 𝑥 + 𝐶 => solution: 2𝑦 + sin(2𝑦) = 4 tan−1 𝑥 + 𝐶

Exact Equations:

Form: 𝜕𝐹

𝜕𝑥𝑑𝑥 +

𝜕𝐹

𝜕𝑦𝑑𝑦 = 0, where

𝜕𝐹

𝜕𝑥= 𝑀,

𝜕𝐹

𝜕𝑦= 𝑁, and 𝐹(𝑥, 𝑦) = 𝐶

The differential equation is exact <=> 𝜕𝑀

𝜕𝑦=

𝜕𝑁

𝜕𝑥

If exact: 1. Integrate 𝜕𝐹

𝜕𝑥 with respect to x or

𝜕𝐹

𝜕𝑦 with respect to y to find 𝐹(𝑥, 𝑦), where constant

𝐶 is a function of the other variable.

2. Find partial of 𝐹(𝑥, 𝑦) with respect to the other variable

Set equal to the original partial in order to find value of 𝐶′

Integrate 𝐶′ to find value of 𝐶

3. Final solution should be in the form 𝐹(𝑥, 𝑦) = 𝐶

Example: (cos 𝑥 cos 𝑦 + 2𝑥)𝑑𝑥 − (sin 𝑥 sin 𝑦 + 2𝑦)𝑑𝑦 = 0

𝜕𝐹

𝜕𝑥= 𝑀 = cos 𝑥 cos 𝑦 + 2𝑥

𝜕𝐹

𝜕𝑦= 𝑁 = −sin 𝑥 sin 𝑦 − 2𝑦

𝜕𝑀

𝜕𝑦= −cos 𝑥 sin 𝑦

𝜕𝑁

𝜕𝑥= −cos 𝑥 sin 𝑦, therefore this is an exact equation

𝐹(𝑥, 𝑦) = ∫𝜕𝐹

𝜕𝑥𝑑𝑥 = ∫(cos 𝑥 cos 𝑦 + 2𝑥)𝑑𝑥 = sin 𝑥 cos 𝑦 + 𝑥2 + 𝐶(𝑦)

𝜕𝐹

𝜕𝑦=

𝜕

𝜕𝑦(sin 𝑥 cos 𝑦 + 𝑥2 + 𝐶(𝑦)) = − sin 𝑥 sin 𝑦 + 𝐶′(𝑦) = − sin 𝑥 sin 𝑦 − 2𝑦

𝐶′(𝑦) = −2𝑦 => 𝐶(𝑦) = −𝑦2 𝐹(𝑥, 𝑦) = sin 𝑥 cos 𝑦 + 𝑥2 − 𝑦2

solution: sin 𝑥 cos 𝑦 + 𝑥2 − 𝑦2 = 𝐶

Bernoulli Equations:

Form: 𝑑𝑦

𝑑𝑥+ 𝑝(𝑥)𝑦 = 𝑞(𝑥)𝑦𝑛

Find solution by substituting 𝑣 = 𝑦1−𝑛 if 𝑦 ≢ 0, and 𝑑𝑣

𝑑𝑥= (1 − 𝑛)𝑦−𝑛 𝑑𝑦

𝑑𝑥

Divide equation through by 𝑦𝑛 then substitute in 𝑣 and 𝑑𝑣

𝑑𝑥

After substituting, the equation is left in a form that is usually linear.

Example: 𝑑𝑦

𝑑𝑥+

𝑦

𝑥= 𝑥2𝑦2 𝑣 = 𝑦1−𝑛 = 𝑦1−2 = 𝑦−1 −

𝑑𝑣

𝑑𝑥= 𝑦−2 𝑑𝑦

𝑑𝑥

=> 𝑦−2 𝑑𝑦

𝑑𝑥+ 𝑥−1𝑦−1 = 𝑥2 => −

𝑑𝑣

𝑑𝑥+

𝑣

𝑥= 𝑥2 =>

𝑑𝑣

𝑑𝑥−

𝑣

𝑥= −𝑥2

µ(𝑥) = 𝑒− ∫𝑑𝑥

𝑥 = 𝑒− ln|𝑥| = 𝑒ln|𝑥−1| = 𝑥−1 =1

𝑥

=> 𝑥−1 𝑑𝑣

𝑑𝑥−

𝑣

𝑥2 = −𝑥 => 𝑑

𝑑𝑥(𝑥−1𝑣) = −𝑥 => 𝑥−1𝑣 = − ∫ 𝑥𝑑𝑥

=> 𝑣

𝑥= −

1

2𝑥2 + 𝐶 => 𝑣 = −

1

2𝑥3 + 𝐶𝑥 = 𝑦−1

Solution: 𝑦 = 2

𝐶𝑥−𝑥3 and 𝑦 ≡ 0

Homogeneous Linear Equations:

Form: 𝑎𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 0

try solution form: 𝑦 = 𝑒𝑟𝑡

change equation to form: 𝑎𝑟2 + 𝑏𝑟 + 𝑐 = 0

then factor or use 𝑟 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎 to solve for zeroes of r

general solution form: 𝑦 = 𝑐1𝑒𝑟1𝑡 + 𝑐2𝑒𝑟2𝑡

If there is a repeated root, solution form: 𝑦 = 𝑐1𝑒𝑟𝑡 + 𝑐2𝑡𝑒𝑟𝑡

If there is a complex root, ex: 𝑟 = 𝑎 ± 𝑏𝑖

Then solution is in the form: 𝑦 = 𝑐1𝑒𝑎𝑡 cos 𝑏𝑡 + 𝑐2𝑒𝑎𝑡 sin 𝑏𝑡

Method of Undetermined Coefficients:

Form: 𝑎𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 𝑓(𝑡)

Guess a solution 𝑦𝑝 that is in a form similar to 𝑓(𝑡)

Ex: 𝑓(𝑡) = 𝑒2𝑡 => 𝑦𝑝 = 𝐴𝑒2𝑡,

𝑓(𝑡) = 𝑡2sin 𝑡 => 𝑦𝑝 = (𝐴𝑡2 + 𝐵𝑡 + 𝐶) sin 𝑡 + (𝐷𝑡2 + 𝐸𝑡 + 𝐹) cos 𝑡

But always solve homogeneous equation first!

If homogeneous solution contains term with a form similar to 𝑓(𝑡), multiply 𝑦𝑝 times 𝑡

Ex: If 𝑓(𝑡) = 2𝑒−𝑡 and 𝑦ℎ = 𝑐1𝑒−𝑡 + 𝑐2𝑡𝑒−𝑡, then guess 𝑦𝑝 = 𝑡2𝐴𝑒−𝑡

After you guess, find 𝑦𝑝′ and 𝑦𝑝

′′ then plug 𝑦𝑝 and its derivatives into the original equation.

Solve for constants so that the left hand side is equal to 𝑓(𝑡)

Plug in constants into original guess 𝑦𝑝

General solution 𝑦 is sum of 𝑦𝑝 and 𝑦ℎ

Example: 𝑦′′ − 5𝑦′ + 6𝑦 = 𝑥𝑒𝑥

𝑦′′ − 5𝑦′ + 6𝑦 = 0 𝑦ℎ = 𝑒𝑟𝑥 => 𝑟2 − 5𝑟 + 6 = 0 => (𝑟 − 3)(𝑟 − 2) = 0 => 𝑟 = 2, 3

𝑦ℎ = 𝑐1𝑒2𝑥 + 𝑐2𝑒3𝑥

𝑦𝑝 = (𝐴𝑥 + 𝐵)𝑒𝑥 = 𝐴𝑥𝑒𝑥 + 𝐵𝑒𝑥 𝑦𝑝′ = 𝐴𝑥𝑒𝑥 + 𝐴𝑒𝑥 + 𝐵𝑒𝑥 𝑦𝑝

′′ = 𝐴𝑥𝑒𝑥 + 2𝐴𝑒𝑥 + 𝐵𝑒𝑥

=> 𝐴𝑥𝑒𝑥 + 2𝐴𝑒𝑥 + 𝐵𝑒𝑥 − 5𝐴𝑥𝑒𝑥 − 5𝐴𝑒𝑥 − 𝐵𝑒𝑥 + 6𝐴𝑥𝑒𝑥 + 6𝐵𝑒𝑥 = 𝑥𝑒𝑥

=> 2𝐴𝑥𝑒𝑥 − 3𝐴𝑒𝑥 + 2𝐵𝑒𝑥 = 𝑥𝑒𝑥 ∴ 𝐴 =1

2 , 𝐵 =

3

4

𝑦𝑝 =1

2𝑥𝑒𝑥 +

3

4𝑒𝑥

General solution: 𝑦 =1

2𝑥𝑒𝑥 +

3

4𝑒𝑥 + 𝑐1𝑒2𝑥 + 𝑐2𝑒3𝑥

Variation of Parameters:

Form: 𝑎(𝑥)𝑦′′ + 𝑏(𝑥)𝑦′ + 𝑐(𝑥)𝑦 = 𝑓(𝑥)

1. Find solutions to the homogeneous equation and denote them as 𝑦1 and 𝑦2

2. Find a solution to the original equation, by guessing 𝑦𝑝 = 𝑣1𝑦1 + 𝑣2𝑦2

Require that: 𝑣1′ 𝑦1 + 𝑣2

′ 𝑦2 = 0 and 𝑣1′ 𝑦1

′ + 𝑣2′ 𝑦2

′ =𝑓(𝑥)

𝑎(𝑥) and use this system of equations

to solve for 𝑣1 and 𝑣2

Example: 𝑦′′ + 𝑦 = 3 sec 𝑡 − 𝑡2 + 1

𝑦′′ + 𝑦 = 0 𝑦ℎ = 𝑒𝑟𝑡 => 𝑟2 + 1 = 0 => 𝑟 = ±𝑖 => 𝑦ℎ = 𝑐1 cos 𝑡 + 𝑐2 sin 𝑡

𝑦1 = cos 𝑡 𝑦2 = sin 𝑡 𝑦𝑝 = 𝑣1 cos 𝑡 + 𝑣2 sin 𝑡

(𝑣1′ cos 𝑡 + 𝑣2

′ sin 𝑡 = 0) sin 𝑡

(𝑣2′ cos 𝑡 − 𝑣1

′ sin 𝑡 = 3 sec 𝑡 − 𝑡2 + 1) cos 𝑡

=> 𝑣2′ sin2 𝑡 + 𝑣2

′ cos2 𝑡 = 3 sec 𝑡 cos 𝑡 − 𝑡2 cos 𝑡 + cos 𝑡

=> 𝑣2′ (sin2 𝑡 + cos2 𝑡) = 3 − 𝑡2 cos 𝑡 + cos 𝑡 => 𝑣2 = ∫ 3𝑑𝑡 − ∫ 𝑡2 cos 𝑡 𝑑𝑡 + ∫ cos 𝑡 𝑑𝑡

=> 𝑣2 = 3𝑡 + sin 𝑡 − ∫ 𝑡2 cos 𝑡 𝑑𝑡 𝑢 = 𝑡2 𝑑𝑢 = 2𝑡𝑑𝑡 𝑑𝑣 = cos 𝑡 𝑑𝑡 𝑣 = sin 𝑡

=> 𝑣2 = 3𝑡 + sin 𝑡 − 𝑡2 sin 𝑡 + ∫ 2𝑡 sin 𝑡 𝑑𝑡 𝑢 = 2𝑡 𝑑𝑢 = 2𝑑𝑡 𝑑𝑣 = sin 𝑡 𝑣 = − cos 𝑡

=> 𝑣2 = 3𝑡 + sin 𝑡 − 𝑡2 sin 𝑡 − 2𝑡 cos 𝑡 + 2 ∫ cos 𝑡 𝑑𝑡 = 3𝑡 − 𝑡2 sin 𝑡 − 2𝑡 cos 𝑡 + 3 sin 𝑡

𝑣1′ cos 𝑡 + (3 − 𝑡2 cos 𝑡 + cos 𝑡) sin 𝑡 = 0 => 𝑣1

′ =−3 sin 𝑡+𝑡2 sin 𝑡 cos 𝑡−cos 𝑡 sin 𝑡

cos 𝑡

=> 𝑣1 = −3 ∫sin 𝑡

cos 𝑡𝑑𝑡 + ∫ 𝑡2 sin 𝑡 𝑑𝑡 − ∫ sin 𝑡 𝑑𝑡

𝑢 = cos 𝑡 − 𝑑𝑢 = sin 𝑡 𝑑𝑡 => 3 ∫𝑑𝑢

𝑢= 3 ln|𝑢| = 3 ln|cos 𝑡|

∫ 𝑡2 sin 𝑡 𝑑𝑡 𝑢 = 𝑡2 𝑑𝑢 = 2𝑡𝑑𝑡 𝑑𝑣 = sin 𝑡 𝑑𝑡 𝑣 = − cos 𝑡

=> −𝑡2 cos 𝑡 + ∫ 2𝑡 cos 𝑡 𝑑𝑡 𝑢 = 2𝑡 𝑑𝑢 = 2𝑑𝑡 𝑑𝑣 = cos 𝑡 𝑑𝑡 𝑣 = sin 𝑡

=> −𝑡2 cos 𝑡 + 2𝑡 sin 𝑡 − 2 ∫ sin 𝑡 𝑑𝑡 = −𝑡2 cos 𝑡 + 2𝑡 sin 𝑡 + 2 cos 𝑡

=> 𝑣1 = 3 ln|cos 𝑡| − 𝑡2 cos 𝑡 + 2𝑡 sin 𝑡 + 2 cos 𝑡 + cos 𝑡

=> 𝑦𝑝 = (3 ln|cos 𝑡| − 𝑡2 cos 𝑡 + 2𝑡 sin 𝑡 + 2 cos 𝑡 + cos 𝑡) cos 𝑡

+(3𝑡 − 𝑡2 sin 𝑡 − 2𝑡 cos 𝑡 + 3 sin 𝑡) sin 𝑡

= 3 (cos 𝑡) ln|cos 𝑡| − 𝑡2 cos2 𝑡 + 2𝑡 sin 𝑡 cos 𝑡 + 3 cos2 𝑡

+3𝑡 sin 𝑡 − 𝑡2 sin2 𝑡 − 2𝑡 sin 𝑡 cos 𝑡 + 3 sin2 𝑡

= 3 (cos 𝑡) ln|cos 𝑡| − 𝑡2 + 3 + 3𝑡 sin 𝑡

General solution: 𝑦 = 𝑐1 cos 𝑡 + 𝑐2 sin 𝑡 + 3 (cos 𝑡) ln|cos 𝑡| − 𝑡2 + 3 + 3𝑡 sin 𝑡

Cauchy-Euler Equation:

Form: 𝑎𝑡2𝑦′′ + 𝑏𝑡𝑦′ + 𝑐𝑦 = 0

Try solution form: 𝑦 = 𝑡𝑟 𝑦′ = 𝑟𝑡𝑟−1 𝑦′′ = 𝑟(𝑟 − 1)𝑡𝑟−2

Substitute 𝑦 and its derivatives into the original equation and solve for zeros of r

Ex: If 𝑟 = −1, 2 𝑦ℎ = 𝑐1𝑡−1 + 𝑐2𝑡2

If you have a repeated root, multiply second solution by ln 𝑡

Ex: If 𝑟 = 3, 3 𝑦ℎ = 𝑐1𝑡3 + 𝑐2 ln(𝑡) 𝑡3

Reduction of Order:

Form: Any second order, linear, homogeneous equations

1. This method is used for equations in which you already have one solution, in order to find

the second solution

2. Guess a second solution such that 𝑦2 = 𝑣𝑦1 where 𝑣 is a function of 𝑡

3. Find 𝑦2′ and 𝑦2

′′ then plug 𝑦2 and its derivatives into the original equation

4. The equation should simplify to an equation containing a 𝑣′ and 𝑣′′ term

5. Substitute 𝑤 = 𝑣′ and 𝑤′ = 𝑣′′ into the equation, reducing the equation to first order

6. Then solve using any possible first order method

7. Once you solve for 𝑤, find 𝑣 by integrating, because 𝑤 = 𝑣′

8. Plug 𝑣 into 𝑦2 to give a second linearly independent solution

Example: 𝑡2𝑦′′ − 2𝑡𝑦′ − 4𝑦 = 0 𝑦1 = 𝑡−1

𝑦2 = 𝑣𝑡−1 𝑦2′ = 𝑣′𝑡−1 − 𝑣𝑡−2 𝑦2

′′ = 𝑣′′𝑡−1 − 2𝑣′𝑡−2+ 2𝑣𝑡−3

𝑡2(𝑣′′𝑡−1 − 2𝑣′𝑡−2 + 2𝑣𝑡−3) − 2𝑡(𝑣′𝑡−1 − 𝑣𝑡−2) − 4(𝑣𝑡−1) = 0

𝑣′′t − 2𝑣′ + 2𝑣𝑡−1 − 2𝑣′ + 2𝑣𝑡−1 − 4𝑣𝑡−1 = 0 => 𝑣′′t − 4𝑣′ = 0

𝑤 = 𝑣′ 𝑤′ = 𝑣′′ => 𝑤′𝑡 − 4𝑤 = 0 => 𝑑𝑤

𝑑𝑡−

4𝑤

𝑡= 0

=> 𝜇(𝑡) = 𝑒− ∫4

𝑡𝑑𝑡 = 𝑒−4 ln|t| = 𝑒ln|t−4| = 𝑡−4

=> 𝑡−4 𝑑𝑤

𝑑𝑡− 4𝑡−3𝑤 = 0 =>

𝑑

𝑑𝑡(𝑡−4𝑤) = 0 => 𝑡−4𝑤 = 𝐶 => 𝑤 = 𝐶𝑡4

=> 𝑣′ = 𝐶𝑡4 => 𝑣 = 𝐶 ∫ 𝑡4𝑑𝑡 =𝐶

5𝑡5 + 𝐷 𝐶 = 5, 𝐷 = 0 𝑣 = 𝑡5

𝑦2 = 𝑡5𝑡−1 = 𝑡4 General solution: 𝑦 = 𝑐1𝑡−1 + 𝑐2𝑡4

Operators:

An operator is a function whose input and output are both functions

Ex: 𝐷[𝑓(t)] =𝑑f

𝑑𝑡 (𝐷2 + 2)[𝑥2 + 𝑥] = 2 + 0 + 2𝑥2 + 2𝑥 = 2𝑥2 + 2𝑥 + 2

You can use operators to solve systems differential equations, by isolating variables

Example: 𝑑x

𝑑𝑡= 4𝑥 − 𝑦

𝑑y

𝑑𝑡= 𝑦 − 2𝑥 𝑥(0) = 1 𝑦(0) = 0

(𝐷 − 4)𝑥 − 𝑦 = 0 (𝐷 − 1)𝑦 + 2𝑥 = 0

=> [(𝐷 − 4)[𝑥] − 𝑦 = 0](−2) [(𝐷 − 1)[𝑦] + 2𝑥 = 0](𝐷 − 4)

=> 2𝑦 + (𝐷 − 4)(𝐷 − 1)[𝑦] + (𝐷 − 4)[2𝑥] − (𝐷 − 4)[2𝑥] = 0

=> (𝐷 − 4)(𝐷 − 1)[𝑦] + 2𝑦 = 0 => (𝐷 − 4)[y′ − y] + 2𝑦 = 0 => y′′ − 5y′ + 6𝑦 = 0

Try 𝑦 = 𝑒𝑟𝑡 => 𝑟2 − 5𝑟 + 6 = 0 => (𝑟 − 3)(𝑟 − 2) = 0 𝑟 = 2, 3

𝑦 = 𝑐1𝑒2𝑡 + 𝑐2𝑒3𝑡 𝑦′ = 2𝑐1𝑒2𝑡 + 3𝑐2𝑒3𝑡

=> 2𝑐1𝑒2𝑡 + 3𝑐2𝑒3𝑡 = 𝑐1𝑒2𝑡 + 𝑐2𝑒3𝑡 − 2𝑥

=> 𝑐1𝑒2𝑡 + 2𝑐2𝑒3𝑡 = −2𝑥 => 𝑥 = −1

2𝑐1𝑒2𝑡 − 𝑐2𝑒3𝑡

𝑦(0) = 0 = 𝑐1 + 𝑐2 𝑥(0) = 0 = −1

2𝑐1 − 𝑐2 𝑐1 = 2, 𝑐2 = −2

𝑥 = 2𝑒3t − 𝑒2𝑡 𝑦 = 2𝑒2𝑡 − 2𝑒3𝑡

Laplace Transforms:

Gives a way of solving a differential equation to one of solving an algebraic equation

You can use them to solve problems that we couldn’t solve before

Laplace Transform table will be provided on the test

Form: 𝐹(𝑠) = ℒ{𝑓(𝑡)} = ∫ 𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡∞

0

Example: ℒ{sin 2𝑡 sin 5𝑡}

sin 𝑎 sin 𝑏 =1

2(cos(𝑎 − 𝑏) − cos(𝑎 + 𝑏))

=> sin 2𝑡 sin 5𝑡 =1

2(cos(2𝑡 − 5𝑡) − cos(2𝑡 + 5𝑡)) =

1

2(cos(−3𝑡) − cos(7𝑡))

cos(−𝑥) = cos 𝑥 => 1

2(cos 3𝑡 − cos 7𝑡)

=> ℒ {1

2(cos 3𝑡 − cos 7𝑡)} =

1

2ℒ{cos 3𝑡 − cos 7𝑡} =

1

2[

𝑠

𝑠2+32 −𝑠

𝑠2+72]

ℒ{sin 2𝑡 sin 5𝑡} =𝑠

2(𝑠2+9)−

𝑠

2(𝑠2+49)

Inverse Laplace Transform: Opposite operation of Laplace transform

ℒ−1{𝐹(𝑠)} = 𝑓(𝑡)

Use partial fraction decomposition or completing the square to put 𝐹(𝑠) in a form which can

easily be transformed back to 𝑓(𝑡)

For completing the square: Given 𝑎𝑠2 + 𝑏𝑠 to complete the square, 𝑐 = (1

2𝑏)2

=> 𝑎𝑠2 + 𝑏𝑠 + 𝑐

For partial fraction decomposition: split the fraction into terms such that each term’s

denominator is a factor of the original fractions denominator. If there is a higher order term,

there must be a term representing each term below that. If there is an irreducible term in the

denominator, the numerator must be one power lower

Ex: If the denominator contains 𝑥3(𝑥2 + 1) => Partial fraction decomp. =𝐴

𝑥3 +𝐵

𝑥2 +𝐶

𝑥+

𝐷𝑥+𝐸

𝑥2+1

Example: ℒ−1 {3𝑠−15

2𝑠2−4𝑠+10}

3𝑠−15

2𝑠2−4𝑠+10=

3(𝑠−5)

2(𝑠2−2𝑠+5) complete the square 𝑠2 − 2𝑠 + 1 = (𝑠 − 1)2 =>

3

2[

(𝑠−1)−4

(𝑠−1)2+4]

=> 3

2[

(𝑠−1)

(𝑠−1)2+4] −

3

2[

4

(𝑠−1)2+4] =

3

2[

(𝑠−1)

(𝑠−1)2+22] − 3 [2

(𝑠−1)2+22]

ℒ−1 {3𝑠−15

2𝑠2−4𝑠+10} = 𝑓(𝑡) =

3

2𝑒𝑡 cos 2𝑡 − 3𝑒𝑡 sin 2𝑡

Example: ℒ−1 {5𝑠2+34𝑠+53

(𝑠+3)2(𝑠+1)}

5𝑠2+34𝑠+53

(𝑠+3)2(𝑠+1)=

𝐴

(𝑠+3)2 +𝐵

𝑠+3+

𝐶

𝑠+1

=> 5𝑠2 + 34𝑠 + 53 = 𝐴(𝑠 + 1) + 𝐵(𝑠 + 3)(𝑠 + 1) + 𝐶(𝑠 + 3)3

𝑠 = −1 24 = 4𝐶 𝐶 = 6

𝑠 = −3 − 4 = −2𝐴 𝐴 = 2

𝑠 = 0 53 = 2 + 3𝐵 + 54 − 3 = 3𝐵 𝐵 = −1

=> 𝐴

(𝑠+3)2 +𝐵

𝑠+3+

𝐶

𝑠+1=

2

(𝑠+3)2 −1

𝑠+3+

6

𝑠+1

ℒ−1 {5𝑠2+34𝑠+53

(𝑠+3)2(𝑠+1)} = 𝑓(𝑡) = 2𝑒−3𝑡𝑡 − 𝑒−3𝑡 + 6𝑒−𝑡

Series Solutions:

Taylor Polynomial Approximation:

Form for Nth degree Taylor polynomial centered at 𝑥0: 𝑃N(𝑥) = ∑𝑓(𝑛)(𝑥0)

𝑛!(𝑥 − 𝑥0)𝑛𝑁

𝑛=0

Example: determine first 3 terms of Taylor polynomial for the IVT

𝑦′ = sin(𝑥 + 𝑦) 𝑦(0) = 0

𝑦 = 𝑦(0) + 𝑦′(0)𝑥 +𝑦′′(0)

2!𝑥2 +

𝑦′′′(0)

3!𝑥3 +

𝑦′𝑣(0)

4!𝑥4 + ⋯ +

𝑦(𝑛)(0)

𝑛!𝑥𝑛

𝑦′(0) = sin(0 + 𝑦(0)) = sin 0 = 0

𝑦′′ = cos(𝑥 + 𝑦) 𝑦′′(0) = cos(0 + 𝑦(0)) = cos 0 = 1

𝑦′′′ = − sin(𝑥 + 𝑦) 𝑦′′′(0) = − sin(0 + 𝑦(0)) = − sin 0 = 0

𝑦(′v) = −cos(𝑥 + 𝑦) 𝑦′𝑣(0) = −cos(0 + 𝑦(0)) = cos 0 = −1 <= Note pattern!

𝑦(𝑥) = 0 + 0 +1

2!𝑥2 + 0 −

1

4!𝑥4 + 0 +

1

6!𝑥6 =

1

2𝑥2 −

1

24𝑥4 +

1

720𝑥6

Power Series:

Form for power series centered at 𝑥0 : 𝑦(𝑥) = ∑ 𝑎𝑛(𝑥 − 𝑥0)𝑛∞𝑛=0

Finding Convergence Set:

1. Apply ratio test to 𝑎𝑛

2. The result of ratio test 𝐿 is greater than |𝑥 − 𝑥0|

3. Ex. If |𝑥 + 2| < 2 the series converges for −4 < 𝑥 < 0

4. Plug bounds for 𝑥 into the original power series, giving 2 new series

5. When stating the convergence set, if series is convergent, use [] for 𝑥 bound and if

divergent, use () for 𝑥 bound

Example: ∑2−𝑛

𝑛+1(𝑥 − 1)𝑛∞

𝑛=0

lim𝑛→∞ |2−𝑛

𝑛+1∙

𝑛+2

2−(𝑛+1)| = lim𝑛→∞ |1

𝑛+1∙

𝑛+2

2−1 | = 2 lim𝑛→∞

|𝑛+2

𝑛+1| = 2

|𝑥 − 1| < 2 ∴ the series converges for −1 < 𝑥 < 3

@ 𝑥 = −1

∑2−𝑛

𝑛+1(−2)𝑛∞

𝑛=0 = ∑2−𝑛

𝑛+1(2)𝑛(−1)𝑛∞

𝑛=0 = ∑(−1)𝑛

𝑛+1∞𝑛=0 alternating harmonic series

Test for absolute convergence ∑ |(−1)𝑛

𝑛+1|∞

𝑛=0 = ∑1

𝑛+1∞𝑛=0 => harmonic series, not abs. convergent

Test for conditional convergence alternating series test: series is decreasing, and alternating

lim𝑛→∞1

𝑛+1= 0 ∴ conditionally convergent at 𝑥 = −1

@ 𝑥 = 3

∑2−𝑛

𝑛+1(2)𝑛∞

𝑛=0 = ∑1

𝑛+1∞𝑛=0 => harmonic series, p-series with p=1, so divergent

The series is divergent at 𝑥 = 3

The convergent set is [-1, 3)

Remember: You can integrate and derive a power series, just like any other function, but you may have

to shift the starting index

Example: 𝑦′(𝑥) = ∑ 𝑛𝑎𝑛(𝑥 − 𝑥0)𝑛−1∞𝑛=1

∫ 𝑦(𝑥)𝑑𝑥 = ∑𝑎𝑛

𝑛+1(𝑥 − 𝑥0)𝑛+1∞

𝑛=0

Using a power series to solve a differential equation:

1. Use initial form: 𝑦(𝑥) = ∑ 𝑎𝑛𝑥𝑛∞𝑛=0 so 𝑦′(𝑥) = ∑ 𝑛𝑎𝑛𝑥𝑛−1∞

𝑛=1

𝑦′′(𝑥) = ∑ 𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2∞𝑛=2

2. Plug 𝑦(𝑥) and its derivatives into the original differential equation.

3. Use a substitution so that 𝑥 contains the same power throughout the function

4. You can then combine the summations, pulling out lower indexed terms, if needed

5. Set any outside terms of the same power of 𝑥 = 0 , and set the summation = 0

6. Solve for highest indexed term of 𝑎𝑛 in the summation so it can be in terms of lower

indexed terms.

7. After solving for each index of 𝑎𝑛 plug them into 𝑦(𝑥)

Example: Find the first 4 non-zero terms in a power series expansion about 𝑥0 = 0

𝑧′′ − 𝑥2𝑧 = 0

𝑧(𝑥) = ∑ 𝑎𝑛𝑥𝑛∞𝑛=0 𝑧′(𝑥) = ∑ 𝑛𝑎𝑛𝑥𝑛−1∞

𝑛=1 𝑧′′(𝑥) = ∑ 𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2∞𝑛=2

∑ 𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2∞𝑛=2 − 𝑥2 ∑ 𝑎𝑛𝑥𝑛∞

𝑛=0 = 0 => ∑ 𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2∞𝑛=2 − ∑ 𝑎𝑛𝑥𝑛+2∞

𝑛=0 = 0

Substitute 𝑘 = 𝑛 − 2 in first summation and 𝑘 = 𝑛 + 2 in second summation to make 𝑥 the

same power in both summations

=> ∑ (𝑘 + 2)(𝑘 + 1)𝑎𝑘+2𝑥𝑘∞𝑘=0 − ∑ 𝑎𝑘−2𝑥𝑘∞

𝑘=2 = 0

=> 2𝑎2 + 6𝑎3𝑥 + ∑ ((𝑘 + 2)(𝑘 + 1)𝑎𝑘+2 − 𝑎𝑘−2)𝑥𝑘∞𝑘=2 = 0

2𝑎2 = 0 6𝑎3 = 0 𝑎2 = 0 𝑎3 = 0

𝑎𝑘+2 =𝑎𝑘−2

(𝑘+2)(k+1)

𝑘 = 2 𝑎4 =𝑎0

12 𝑘 = 3 𝑎5 =

𝑎1

20

=> 𝑧(𝑥) = 𝑎0 + 𝑎1𝑥 + 0 + 0 + 𝑎4𝑥4 + 𝑎5𝑥5 = 𝑎0 + 𝑎1𝑥 +1

12𝑎0𝑥4 +

1

20𝑎1𝑥5

= 𝑎0 (1 +1

12𝑥4 … ) + 𝑎1 (𝑥 +

1

20𝑥5 … ) + ⋯