In this poblem wewill derive the loinomial thaorem.3. a)use separaton of variablas to solve the ihrtial value probleen-(4X)y-ay0) Find the paver senes solution of theinitial vaie probume l11(tXIy yC) Explain why this shows(Itx +axt ala-x (a-1(a-2)+171x13!2 1d) Show that ifne Nthis giveswhere20n!k! (n-K)

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Asked Oct 10, 2019
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In this poblem wewill derive the loinomial thaorem.
3. a)use separaton of variablas to solve the ihrtial value probleen-
(4X)y-ay
0) Find the paver senes solution of the
initial vaie probum
e l
11
(tXIy y
C) Explain why this shows
(Itx +axt ala-x (a-1(a-2)
+
171x1
3!
2 1
d) Show that if
ne Nthis gives
where
20
n!
k! (n-K)
help_outline

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In this poblem wewill derive the loinomial thaorem. 3. a)use separaton of variablas to solve the ihrtial value probleen- (4X)y-ay 0) Find the paver senes solution of the initial vaie probum e l 11 (tXIy y C) Explain why this shows (Itx +axt ala-x (a-1(a-2) + 171x1 3! 2 1 d) Show that if ne Nthis gives where 20 n! k! (n-K)

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Expert Answer

Step 1

3 a]

The first order differential equation is given as (1+x) y' = ay with initial value y(0) =1
Use separation ofvariables, solve the differential equations
dy
(1+x)
ay
dx
dy
dc
1+x
In yl= aIn|1+x|+C
y= ep+zf°+c_
y= 4(1+x)
:A= ec
Thus, the solution becomes
y(0)4(1-(0)
A = 1
y =(1+x)®
Therefore, the first order differential equation is y = (1+x)
help_outline

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The first order differential equation is given as (1+x) y' = ay with initial value y(0) =1 Use separation ofvariables, solve the differential equations dy (1+x) ay dx dy dc 1+x In yl= aIn|1+x|+C y= ep+zf°+c_ y= 4(1+x) :A= ec Thus, the solution becomes y(0)4(1-(0) A = 1 y =(1+x)® Therefore, the first order differential equation is y = (1+x)

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Step 2

3 b]

Consider the solution of the first order differential equation y =a^x* .
k-0
The first order derivative is y' = kax'
k=1
Substitute y andy' in first order differential equation and obtain the result
(1+ x ka^x*
k1
+
=
k-0
Σαr-(& - α)Σ α' -
+
kl
k-0
(a, 2a,x+3a,x+ 4ax +..) +(-aa, +(1- a)ax+(2-a)a,x +(3-a)a,x'...)= o
That is
a-aa 0; 2a +(1-a)a = 0;3a, + (2 - a)a, = 0;..
-(1-a)a_(a-1) a
(α-2)(α-1)α,
α- αα,-αα
aa a2
2
2
3-2
help_outline

Image Transcriptionclose

Consider the solution of the first order differential equation y =a^x* . k-0 The first order derivative is y' = kax' k=1 Substitute y andy' in first order differential equation and obtain the result (1+ x ka^x* k1 + = k-0 Σαr-(& - α)Σ α' - + kl k-0 (a, 2a,x+3a,x+ 4ax +..) +(-aa, +(1- a)ax+(2-a)a,x +(3-a)a,x'...)= o That is a-aa 0; 2a +(1-a)a = 0;3a, + (2 - a)a, = 0;.. -(1-a)a_(a-1) a (α-2)(α-1)α, α- αα,-αα aa a2 2 2 3-2

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Step 3

Thus,

...
The solution of (1x)y' = ay becomes
y = ao+a,x + azx2+a,x'++a.x*-
+...
α(α-1)- α(α-1) (α-2),
=1+ax
2
3-2
α(α-1) (α-2) (α-3) ,
..
4 3.2
α(α-1),, α(α-1)(α-2),
=1+ax
2!
3!
α (α-1) (α- 2)(α- 3)
...
4!
Therefore, the power series solution of the first order differential equation is
α(α-1) (α-2)
α (α-1)
α (α-1) (α-2) (α-3)
y 1ax
2!
3!
4!
help_outline

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The solution of (1x)y' = ay becomes y = ao+a,x + azx2+a,x'++a.x*- +... α(α-1)- α(α-1) (α-2), =1+ax 2 3-2 α(α-1) (α-2) (α-3) , .. 4 3.2 α(α-1),, α(α-1)(α-2), =1+ax 2! 3! α (α-1) (α- 2)(α- 3) ... 4! Therefore, the power series solution of the first order differential equation is α(α-1) (α-2) α (α-1) α (α-1) (α-2) (α-3) y 1ax 2! 3! 4!

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