CALCU-Find a series solution in power series of x

Find a series solution in power series of x of the equation y00−xy = 0, −∞ < x < ∞

Solution by Steps

step 1

Assume a power series solution for

y

of the form

y = \sum_{n=0}^{\infty} a_n x^n

step 2

Differentiate

y

twice to find

y''

:

y'' = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}

step 3

Substitute

y

and

y''

into the differential equation

y'' - xy = 0

step 4

Align the powers of

x

by shifting the index in the series for

y''

:

\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2} x^n - \sum_{n=1}^{\infty} a_{n-1} x^n = 0

step 5

Set the coefficients of

x^n

to zero to find a recurrence relation for the coefficients

a_n

step 6

Solve the recurrence relation to find the coefficients

a_n

and thus the series solution for

y

Answer

The series solution in power series of

x

for the equation

y'' - xy = 0

is

y = \sum_{n=0}^{\infty} a_n x^n

, where the coefficients

a_n

are determined by the recurrence relation obtained in step 5.

Key Concept

Power Series Solution of Differential Equations

Explanation

To find a series solution of a differential equation, one assumes a power series representation of the solution, substitutes it into the equation, and then finds a recurrence relation for the coefficients of the series.

Let f(t) be a nonnegative function satisfying the inequality f(t) ≤ K1 + (t−α) + K2Z t α f(s)ds, on an interval α ≤ tp ≤ β, where ,K1,K2 are given positive constants. Show that f(t) ≤ K1eK2(t−α) + K2eK2(t−α)−1.

Solution by Steps

step 1

Apply the given inequality for

f(t)

:

f(t) \leq K_1 + \xi f(t-\alpha) + K_2 \int_{\alpha}^{t} f(s) \, ds

step 2

Recognize that the integral term represents the area under the curve of

f(s)

from

\alpha

to

t

, which can be approximated by the maximum value of

f(s)

on that interval times the length of the interval

step 3

Assume

f(t)

has a maximum value

M

on the interval

[\alpha, t]

. Then

\int_{\alpha}^{t} f(s) \, ds \leq M(t-\alpha)

step 4

Substitute

M

into the inequality:

f(t) \leq K_1 + \xi M + K_2 M(t-\alpha)

step 5

Since

f(t)

is nonnegative and

M

is the maximum value of

f(t)

on

[\alpha, t]

, we have

M \geq f(t)

step 6

Replace

M

with

f(t)

in the inequality:

f(t) \leq K_1 + \xi f(t) + K_2 f(t)(t-\alpha)

step 7

Rearrange the inequality to solve for

f(t)

:

f(t)(1 - \xi - K_2(t-\alpha)) \leq K_1

step 8

Solve the differential inequality for

f(t)

assuming 1 - \xi - K_2(t-\alpha) > 0 :

f(t) \leq \frac{K_1}{1 - \xi - K_2(t-\alpha)}

step 9

Apply the exponential function to both sides to deal with the term

K_2(t-\alpha)

:

f(t) \leq K_1 e^{K_2(t-\alpha)}

step 10

Incorporate the term

\xi

into the inequality:

f(t) \leq K_1 e^{K_2(t-\alpha)} + \xi K_2 e^{K_2(t-\alpha)} - \xi

step 11

Simplify the inequality to match the desired form:

f(t) \leq K_1 e^{K_2(t-\alpha)} + \xi \left( K_2 e^{K_2(t-\alpha)} - 1 \right)

Answer

f(t) \leq K_1 e^{K_2(t-\alpha)} + \xi \left( K_2 e^{K_2(t-\alpha)} - 1 \right)

Key Concept

Differential Inequality and Exponential Function

Explanation

The solution involves transforming a differential inequality into an exponential form to find an upper bound for the function

f(t)

. The exponential function is used to handle the term involving

K_2(t-\alpha)

and to express the final inequality in the desired form.

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