differential equations i - study guide
DESCRIPTION
Study Guide for differential equations ITRANSCRIPT
Differential Equations I β Final Exam Review
Linear Equations:
Form: π1(π₯)ππ¦
ππ₯= π0(π₯)π¦ = π(π₯)
Standard form : ππ¦
ππ₯+ π(π₯)π¦ = π(π₯) , where π(π₯) =
π0(π₯)
π1(π₯) and π(π₯) =
π(π₯)
π1(π₯)
If in standard form: multiply through by Β΅(π₯) = πβ« π(π₯)ππ₯
Example: π₯ππ¦
ππ₯+ 2π¦ = π₯β3
ππ¦
ππ₯+
2π¦
π₯= π₯β4 => Β΅(π₯) = πβ«
2
π₯ππ₯ = π2 ln|π₯| = πln|π₯2| = π₯2
=> π₯2 ππ¦
ππ₯+ 2π₯π¦ = π₯β2 =>
π
ππ₯(π₯2y) = π₯β2 => π₯2π¦ = β« π₯β2ππ₯ => π₯2π¦ = βπ₯β1 + πΆ
=> solution: π¦ = βπ₯β3 + πΆπ₯β2 = πΆ
π₯2 β1
π₯3
Separable Equations:
Form: ππ¦
ππ₯= π(π¦)π(π₯)
Get all y dependence on one side and x dependence on the other, then integrate and solve for y.
Example: ππ¦
ππ₯=
sec2 π¦
1+π₯2
ππ¦
sec2 π¦=
ππ₯
1+π₯2 => cos2 π¦ ππ¦ =ππ₯
1+π₯2 => β« cos2 π¦ ππ¦ = β«1
1+π₯2 ππ₯ => β«1+cos 2π¦
2ππ¦ = β«
1
1+π₯2 ππ₯
=> 1
2π¦ +
1
4sin(2π¦) = tanβ1 π₯ + πΆ => solution: 2π¦ + sin(2π¦) = 4 tanβ1 π₯ + πΆ
Exact Equations:
Form: ππΉ
ππ₯ππ₯ +
ππΉ
ππ¦ππ¦ = 0, where
ππΉ
ππ₯= π,
ππΉ
ππ¦= π, and πΉ(π₯, π¦) = πΆ
The differential equation is exact <=> ππ
ππ¦=
ππ
ππ₯
If exact: 1. Integrate ππΉ
ππ₯ with respect to x or
ππΉ
ππ¦ with respect to y to find πΉ(π₯, π¦), where constant
πΆ is a function of the other variable.
2. Find partial of πΉ(π₯, π¦) with respect to the other variable
Set equal to the original partial in order to find value of πΆβ²
Integrate πΆβ² to find value of πΆ
3. Final solution should be in the form πΉ(π₯, π¦) = πΆ
Example: (cos π₯ cos π¦ + 2π₯)ππ₯ β (sin π₯ sin π¦ + 2π¦)ππ¦ = 0
ππΉ
ππ₯= π = cos π₯ cos π¦ + 2π₯
ππΉ
ππ¦= π = βsin π₯ sin π¦ β 2π¦
ππ
ππ¦= βcos π₯ sin π¦
ππ
ππ₯= βcos π₯ sin π¦, therefore this is an exact equation
πΉ(π₯, π¦) = β«ππΉ
ππ₯ππ₯ = β«(cos π₯ cos π¦ + 2π₯)ππ₯ = sin π₯ cos π¦ + π₯2 + πΆ(π¦)
ππΉ
ππ¦=
π
ππ¦(sin π₯ cos π¦ + π₯2 + πΆ(π¦)) = β sin π₯ sin π¦ + πΆβ²(π¦) = β sin π₯ sin π¦ β 2π¦
πΆβ²(π¦) = β2π¦ => πΆ(π¦) = βπ¦2 πΉ(π₯, π¦) = sin π₯ cos π¦ + π₯2 β π¦2
solution: sin π₯ cos π¦ + π₯2 β π¦2 = πΆ
Bernoulli Equations:
Form: ππ¦
ππ₯+ π(π₯)π¦ = π(π₯)π¦π
Find solution by substituting π£ = π¦1βπ if π¦ β’ 0, and ππ£
ππ₯= (1 β π)π¦βπ ππ¦
ππ₯
Divide equation through by π¦π then substitute in π£ and ππ£
ππ₯
After substituting, the equation is left in a form that is usually linear.
Example: ππ¦
ππ₯+
π¦
π₯= π₯2π¦2 π£ = π¦1βπ = π¦1β2 = π¦β1 β
ππ£
ππ₯= π¦β2 ππ¦
ππ₯
=> π¦β2 ππ¦
ππ₯+ π₯β1π¦β1 = π₯2 => β
ππ£
ππ₯+
π£
π₯= π₯2 =>
ππ£
ππ₯β
π£
π₯= βπ₯2
Β΅(π₯) = πβ β«ππ₯
π₯ = πβ ln|π₯| = πln|π₯β1| = π₯β1 =1
π₯
=> π₯β1 ππ£
ππ₯β
π£
π₯2 = βπ₯ => π
ππ₯(π₯β1π£) = βπ₯ => π₯β1π£ = β β« π₯ππ₯
=> π£
π₯= β
1
2π₯2 + πΆ => π£ = β
1
2π₯3 + πΆπ₯ = π¦β1
Solution: π¦ = 2
πΆπ₯βπ₯3 and π¦ β‘ 0
Homogeneous Linear Equations:
Form: ππ¦β²β² + ππ¦β² + ππ¦ = 0
try solution form: π¦ = πππ‘
change equation to form: ππ2 + ππ + π = 0
then factor or use π =βπΒ±βπ2β4ππ
2π to solve for zeroes of r
general solution form: π¦ = π1ππ1π‘ + π2ππ2π‘
If there is a repeated root, solution form: π¦ = π1πππ‘ + π2π‘πππ‘
If there is a complex root, ex: π = π Β± ππ
Then solution is in the form: π¦ = π1πππ‘ cos ππ‘ + π2πππ‘ sin ππ‘
Method of Undetermined Coefficients:
Form: ππ¦β²β² + ππ¦β² + ππ¦ = π(π‘)
Guess a solution π¦π that is in a form similar to π(π‘)
Ex: π(π‘) = π2π‘ => π¦π = π΄π2π‘,
π(π‘) = π‘2sin π‘ => π¦π = (π΄π‘2 + π΅π‘ + πΆ) sin π‘ + (π·π‘2 + πΈπ‘ + πΉ) cos π‘
But always solve homogeneous equation first!
If homogeneous solution contains term with a form similar to π(π‘), multiply π¦π times π‘
Ex: If π(π‘) = 2πβπ‘ and π¦β = π1πβπ‘ + π2π‘πβπ‘, then guess π¦π = π‘2π΄πβπ‘
After you guess, find π¦πβ² and π¦π
β²β² then plug π¦π and its derivatives into the original equation.
Solve for constants so that the left hand side is equal to π(π‘)
Plug in constants into original guess π¦π
General solution π¦ is sum of π¦π and π¦β
Example: π¦β²β² β 5π¦β² + 6π¦ = π₯ππ₯
π¦β²β² β 5π¦β² + 6π¦ = 0 π¦β = πππ₯ => π2 β 5π + 6 = 0 => (π β 3)(π β 2) = 0 => π = 2, 3
π¦β = π1π2π₯ + π2π3π₯
π¦π = (π΄π₯ + π΅)ππ₯ = π΄π₯ππ₯ + π΅ππ₯ π¦πβ² = π΄π₯ππ₯ + π΄ππ₯ + π΅ππ₯ π¦π
β²β² = π΄π₯ππ₯ + 2π΄ππ₯ + π΅ππ₯
=> π΄π₯ππ₯ + 2π΄ππ₯ + π΅ππ₯ β 5π΄π₯ππ₯ β 5π΄ππ₯ β π΅ππ₯ + 6π΄π₯ππ₯ + 6π΅ππ₯ = π₯ππ₯
=> 2π΄π₯ππ₯ β 3π΄ππ₯ + 2π΅ππ₯ = π₯ππ₯ β΄ π΄ =1
2 , π΅ =
3
4
π¦π =1
2π₯ππ₯ +
3
4ππ₯
General solution: π¦ =1
2π₯ππ₯ +
3
4ππ₯ + π1π2π₯ + π2π3π₯
Variation of Parameters:
Form: π(π₯)π¦β²β² + π(π₯)π¦β² + π(π₯)π¦ = π(π₯)
1. Find solutions to the homogeneous equation and denote them as π¦1 and π¦2
2. Find a solution to the original equation, by guessing π¦π = π£1π¦1 + π£2π¦2
Require that: π£1β² π¦1 + π£2
β² π¦2 = 0 and π£1β² π¦1
β² + π£2β² π¦2
β² =π(π₯)
π(π₯) and use this system of equations
to solve for π£1 and π£2
Example: π¦β²β² + π¦ = 3 sec π‘ β π‘2 + 1
π¦β²β² + π¦ = 0 π¦β = πππ‘ => π2 + 1 = 0 => π = Β±π => π¦β = π1 cos π‘ + π2 sin π‘
π¦1 = cos π‘ π¦2 = sin π‘ π¦π = π£1 cos π‘ + π£2 sin π‘
(π£1β² cos π‘ + π£2
β² sin π‘ = 0) sin π‘
(π£2β² cos π‘ β π£1
β² sin π‘ = 3 sec π‘ β π‘2 + 1) cos π‘
=> π£2β² sin2 π‘ + π£2
β² cos2 π‘ = 3 sec π‘ cos π‘ β π‘2 cos π‘ + cos π‘
=> π£2β² (sin2 π‘ + cos2 π‘) = 3 β π‘2 cos π‘ + cos π‘ => π£2 = β« 3ππ‘ β β« π‘2 cos π‘ ππ‘ + β« cos π‘ ππ‘
=> π£2 = 3π‘ + sin π‘ β β« π‘2 cos π‘ ππ‘ π’ = π‘2 ππ’ = 2π‘ππ‘ ππ£ = cos π‘ ππ‘ π£ = sin π‘
=> π£2 = 3π‘ + sin π‘ β π‘2 sin π‘ + β« 2π‘ sin π‘ ππ‘ π’ = 2π‘ ππ’ = 2ππ‘ ππ£ = sin π‘ π£ = β cos π‘
=> π£2 = 3π‘ + sin π‘ β π‘2 sin π‘ β 2π‘ cos π‘ + 2 β« cos π‘ ππ‘ = 3π‘ β π‘2 sin π‘ β 2π‘ cos π‘ + 3 sin π‘
π£1β² cos π‘ + (3 β π‘2 cos π‘ + cos π‘) sin π‘ = 0 => π£1
β² =β3 sin π‘+π‘2 sin π‘ cos π‘βcos π‘ sin π‘
cos π‘
=> π£1 = β3 β«sin π‘
cos π‘ππ‘ + β« π‘2 sin π‘ ππ‘ β β« sin π‘ ππ‘
π’ = cos π‘ β ππ’ = sin π‘ ππ‘ => 3 β«ππ’
π’= 3 ln|π’| = 3 ln|cos π‘|
β« π‘2 sin π‘ ππ‘ π’ = π‘2 ππ’ = 2π‘ππ‘ ππ£ = sin π‘ ππ‘ π£ = β cos π‘
=> βπ‘2 cos π‘ + β« 2π‘ cos π‘ ππ‘ π’ = 2π‘ ππ’ = 2ππ‘ ππ£ = cos π‘ ππ‘ π£ = sin π‘
=> βπ‘2 cos π‘ + 2π‘ sin π‘ β 2 β« sin π‘ ππ‘ = βπ‘2 cos π‘ + 2π‘ sin π‘ + 2 cos π‘
=> π£1 = 3 ln|cos π‘| β π‘2 cos π‘ + 2π‘ sin π‘ + 2 cos π‘ + cos π‘
=> π¦π = (3 ln|cos π‘| β π‘2 cos π‘ + 2π‘ sin π‘ + 2 cos π‘ + cos π‘) cos π‘
+(3π‘ β π‘2 sin π‘ β 2π‘ cos π‘ + 3 sin π‘) sin π‘
= 3 (cos π‘) ln|cos π‘| β π‘2 cos2 π‘ + 2π‘ sin π‘ cos π‘ + 3 cos2 π‘
+3π‘ sin π‘ β π‘2 sin2 π‘ β 2π‘ sin π‘ cos π‘ + 3 sin2 π‘
= 3 (cos π‘) ln|cos π‘| β π‘2 + 3 + 3π‘ sin π‘
General solution: π¦ = π1 cos π‘ + π2 sin π‘ + 3 (cos π‘) ln|cos π‘| β π‘2 + 3 + 3π‘ sin π‘
Cauchy-Euler Equation:
Form: ππ‘2π¦β²β² + ππ‘π¦β² + ππ¦ = 0
Try solution form: π¦ = π‘π π¦β² = ππ‘πβ1 π¦β²β² = π(π β 1)π‘πβ2
Substitute π¦ and its derivatives into the original equation and solve for zeros of r
Ex: If π = β1, 2 π¦β = π1π‘β1 + π2π‘2
If you have a repeated root, multiply second solution by ln π‘
Ex: If π = 3, 3 π¦β = π1π‘3 + π2 ln(π‘) π‘3
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 π¦2 = π£π¦1 where π£ is a function of π‘
3. Find π¦2β² and π¦2
β²β² then plug π¦2 and its derivatives into the original equation
4. The equation should simplify to an equation containing a π£β² and π£β²β² term
5. Substitute π€ = π£β² and π€β² = π£β²β² into the equation, reducing the equation to first order
6. Then solve using any possible first order method
7. Once you solve for π€, find π£ by integrating, because π€ = π£β²
8. Plug π£ into π¦2 to give a second linearly independent solution
Example: π‘2π¦β²β² β 2π‘π¦β² β 4π¦ = 0 π¦1 = π‘β1
π¦2 = π£π‘β1 π¦2β² = π£β²π‘β1 β π£π‘β2 π¦2
β²β² = π£β²β²π‘β1 β 2π£β²π‘β2+ 2π£π‘β3
π‘2(π£β²β²π‘β1 β 2π£β²π‘β2 + 2π£π‘β3) β 2π‘(π£β²π‘β1 β π£π‘β2) β 4(π£π‘β1) = 0
π£β²β²t β 2π£β² + 2π£π‘β1 β 2π£β² + 2π£π‘β1 β 4π£π‘β1 = 0 => π£β²β²t β 4π£β² = 0
π€ = π£β² π€β² = π£β²β² => π€β²π‘ β 4π€ = 0 => ππ€
ππ‘β
4π€
π‘= 0
=> π(π‘) = πβ β«4
π‘ππ‘ = πβ4 ln|t| = πln|tβ4| = π‘β4
=> π‘β4 ππ€
ππ‘β 4π‘β3π€ = 0 =>
π
ππ‘(π‘β4π€) = 0 => π‘β4π€ = πΆ => π€ = πΆπ‘4
=> π£β² = πΆπ‘4 => π£ = πΆ β« π‘4ππ‘ =πΆ
5π‘5 + π· πΆ = 5, π· = 0 π£ = π‘5
π¦2 = π‘5π‘β1 = π‘4 General solution: π¦ = π1π‘β1 + π2π‘4
Operators:
An operator is a function whose input and output are both functions
Ex: π·[π(t)] =πf
ππ‘ (π·2 + 2)[π₯2 + π₯] = 2 + 0 + 2π₯2 + 2π₯ = 2π₯2 + 2π₯ + 2
You can use operators to solve systems differential equations, by isolating variables
Example: πx
ππ‘= 4π₯ β π¦
πy
ππ‘= π¦ β 2π₯ π₯(0) = 1 π¦(0) = 0
(π· β 4)π₯ β π¦ = 0 (π· β 1)π¦ + 2π₯ = 0
=> [(π· β 4)[π₯] β π¦ = 0](β2) [(π· β 1)[π¦] + 2π₯ = 0](π· β 4)
=> 2π¦ + (π· β 4)(π· β 1)[π¦] + (π· β 4)[2π₯] β (π· β 4)[2π₯] = 0
=> (π· β 4)(π· β 1)[π¦] + 2π¦ = 0 => (π· β 4)[yβ² β y] + 2π¦ = 0 => yβ²β² β 5yβ² + 6π¦ = 0
Try π¦ = πππ‘ => π2 β 5π + 6 = 0 => (π β 3)(π β 2) = 0 π = 2, 3
π¦ = π1π2π‘ + π2π3π‘ π¦β² = 2π1π2π‘ + 3π2π3π‘
=> 2π1π2π‘ + 3π2π3π‘ = π1π2π‘ + π2π3π‘ β 2π₯
=> π1π2π‘ + 2π2π3π‘ = β2π₯ => π₯ = β1
2π1π2π‘ β π2π3π‘
π¦(0) = 0 = π1 + π2 π₯(0) = 0 = β1
2π1 β π2 π1 = 2, π2 = β2
π₯ = 2π3t β π2π‘ π¦ = 2π2π‘ β 2π3π‘
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 couldnβt solve before
Laplace Transform table will be provided on the test
Form: πΉ(π ) = β{π(π‘)} = β« πβπ π‘π(π‘)ππ‘β
0
Example: β{sin 2π‘ sin 5π‘}
sin π sin π =1
2(cos(π β π) β cos(π + π))
=> sin 2π‘ sin 5π‘ =1
2(cos(2π‘ β 5π‘) β cos(2π‘ + 5π‘)) =
1
2(cos(β3π‘) β cos(7π‘))
cos(βπ₯) = cos π₯ => 1
2(cos 3π‘ β cos 7π‘)
=> β {1
2(cos 3π‘ β cos 7π‘)} =
1
2β{cos 3π‘ β cos 7π‘} =
1
2[
π
π 2+32 βπ
π 2+72]
β{sin 2π‘ sin 5π‘} =π
2(π 2+9)β
π
2(π 2+49)
Inverse Laplace Transform: Opposite operation of Laplace transform
ββ1{πΉ(π )} = π(π‘)
Use partial fraction decomposition or completing the square to put πΉ(π ) in a form which can
easily be transformed back to π(π‘)
For completing the square: Given ππ 2 + ππ to complete the square, π = (1
2π)2
=> ππ 2 + ππ + π
For partial fraction decomposition: split the fraction into terms such that each termβs
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 π₯3(π₯2 + 1) => Partial fraction decomp. =π΄
π₯3 +π΅
π₯2 +πΆ
π₯+
π·π₯+πΈ
π₯2+1
Example: ββ1 {3π β15
2π 2β4π +10}
3π β15
2π 2β4π +10=
3(π β5)
2(π 2β2π +5) complete the square π 2 β 2π + 1 = (π β 1)2 =>
3
2[
(π β1)β4
(π β1)2+4]
=> 3
2[
(π β1)
(π β1)2+4] β
3
2[
4
(π β1)2+4] =
3
2[
(π β1)
(π β1)2+22] β 3 [2
(π β1)2+22]
ββ1 {3π β15
2π 2β4π +10} = π(π‘) =
3
2ππ‘ cos 2π‘ β 3ππ‘ sin 2π‘
Example: ββ1 {5π 2+34π +53
(π +3)2(π +1)}
5π 2+34π +53
(π +3)2(π +1)=
π΄
(π +3)2 +π΅
π +3+
πΆ
π +1
=> 5π 2 + 34π + 53 = π΄(π + 1) + π΅(π + 3)(π + 1) + πΆ(π + 3)3
π = β1 24 = 4πΆ πΆ = 6
π = β3 β 4 = β2π΄ π΄ = 2
π = 0 53 = 2 + 3π΅ + 54 β 3 = 3π΅ π΅ = β1
=> π΄
(π +3)2 +π΅
π +3+
πΆ
π +1=
2
(π +3)2 β1
π +3+
6
π +1
ββ1 {5π 2+34π +53
(π +3)2(π +1)} = π(π‘) = 2πβ3π‘π‘ β πβ3π‘ + 6πβπ‘
Series Solutions:
Taylor Polynomial Approximation:
Form for Nth degree Taylor polynomial centered at π₯0: πN(π₯) = βπ(π)(π₯0)
π!(π₯ β π₯0)ππ
π=0
Example: determine first 3 terms of Taylor polynomial for the IVT
π¦β² = sin(π₯ + π¦) π¦(0) = 0
π¦ = π¦(0) + π¦β²(0)π₯ +π¦β²β²(0)
2!π₯2 +
π¦β²β²β²(0)
3!π₯3 +
π¦β²π£(0)
4!π₯4 + β― +
π¦(π)(0)
π!π₯π
π¦β²(0) = sin(0 + π¦(0)) = sin 0 = 0
π¦β²β² = cos(π₯ + π¦) π¦β²β²(0) = cos(0 + π¦(0)) = cos 0 = 1
π¦β²β²β² = β sin(π₯ + π¦) π¦β²β²β²(0) = β sin(0 + π¦(0)) = β sin 0 = 0
π¦(β²v) = βcos(π₯ + π¦) π¦β²π£(0) = βcos(0 + π¦(0)) = cos 0 = β1 <= Note pattern!
π¦(π₯) = 0 + 0 +1
2!π₯2 + 0 β
1
4!π₯4 + 0 +
1
6!π₯6 =
1
2π₯2 β
1
24π₯4 +
1
720π₯6
Power Series:
Form for power series centered at π₯0 : π¦(π₯) = β ππ(π₯ β π₯0)πβπ=0
Finding Convergence Set:
1. Apply ratio test to ππ
2. The result of ratio test πΏ is greater than |π₯ β π₯0|
3. Ex. If |π₯ + 2| < 2 the series converges for β4 < π₯ < 0
4. Plug bounds for π₯ into the original power series, giving 2 new series
5. When stating the convergence set, if series is convergent, use [] for π₯ bound and if
divergent, use () for π₯ bound
Example: β2βπ
π+1(π₯ β 1)πβ
π=0
limπββ |2βπ
π+1β
π+2
2β(π+1)| = limπββ |1
π+1β
π+2
2β1 | = 2 limπββ
|π+2
π+1| = 2
|π₯ β 1| < 2 β΄ the series converges for β1 < π₯ < 3
@ π₯ = β1
β2βπ
π+1(β2)πβ
π=0 = β2βπ
π+1(2)π(β1)πβ
π=0 = β(β1)π
π+1βπ=0 alternating harmonic series
Test for absolute convergence β |(β1)π
π+1|β
π=0 = β1
π+1βπ=0 => harmonic series, not abs. convergent
Test for conditional convergence alternating series test: series is decreasing, and alternating
limπββ1
π+1= 0 β΄ conditionally convergent at π₯ = β1
@ π₯ = 3
β2βπ
π+1(2)πβ
π=0 = β1
π+1βπ=0 => harmonic series, p-series with p=1, so divergent
The series is divergent at π₯ = 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: π¦β²(π₯) = β πππ(π₯ β π₯0)πβ1βπ=1
β« π¦(π₯)ππ₯ = βππ
π+1(π₯ β π₯0)π+1β
π=0
Using a power series to solve a differential equation:
1. Use initial form: π¦(π₯) = β πππ₯πβπ=0 so π¦β²(π₯) = β ππππ₯πβ1β
π=1
π¦β²β²(π₯) = β π(π β 1)πππ₯πβ2βπ=2
2. Plug π¦(π₯) and its derivatives into the original differential equation.
3. Use a substitution so that π₯ 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 π₯ = 0 , and set the summation = 0
6. Solve for highest indexed term of ππ in the summation so it can be in terms of lower
indexed terms.
7. After solving for each index of ππ plug them into π¦(π₯)
Example: Find the first 4 non-zero terms in a power series expansion about π₯0 = 0
π§β²β² β π₯2π§ = 0
π§(π₯) = β πππ₯πβπ=0 π§β²(π₯) = β ππππ₯πβ1β
π=1 π§β²β²(π₯) = β π(π β 1)πππ₯πβ2βπ=2
β π(π β 1)πππ₯πβ2βπ=2 β π₯2 β πππ₯πβ
π=0 = 0 => β π(π β 1)πππ₯πβ2βπ=2 β β πππ₯π+2β
π=0 = 0
Substitute π = π β 2 in first summation and π = π + 2 in second summation to make π₯ the
same power in both summations
=> β (π + 2)(π + 1)ππ+2π₯πβπ=0 β β ππβ2π₯πβ
π=2 = 0
=> 2π2 + 6π3π₯ + β ((π + 2)(π + 1)ππ+2 β ππβ2)π₯πβπ=2 = 0
2π2 = 0 6π3 = 0 π2 = 0 π3 = 0
ππ+2 =ππβ2
(π+2)(k+1)
π = 2 π4 =π0
12 π = 3 π5 =
π1
20
=> π§(π₯) = π0 + π1π₯ + 0 + 0 + π4π₯4 + π5π₯5 = π0 + π1π₯ +1
12π0π₯4 +
1
20π1π₯5
= π0 (1 +1
12π₯4 β¦ ) + π1 (π₯ +
1
20π₯5 β¦ ) + β―