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.

Consider the integral equation y(t) = y0 + z0t +Z t 0 (t−s)g(s,y(s))ds, y(0) = 1, (1) where g(t,y) and ∂g(t,y) ∂y are continuous in a rectangle R = {(t,y)||t|≤ a, |y−y0|≤ b}. Let |g(t,y)|≤ M, ∂g(t,y) ∂y ≤ K for all (t,y) ∈ R, deﬁneφ 0(t) = y0, φn(t) = y0 + z0t +Z t 0 (t−s)g(s,φn−1(s))ds, n = 1,2,3,··· , Show that (a) φn(t) are well deﬁned for |t|≤ α, where α = mina, b ¯ M, ¯ M = |z0|+ Ma 2 . (b) {φn(t)} converges to a solution of the intehgral equation (1) on |t|≤ α

Solution by Steps

step 1

Define the sequence of functions

\{\varphi_n(t)\}

recursively where

\varphi_0(t) = y_0

and

\varphi_n(t) = y_0 + z_0t + \int_0^t (t-s)g(s,\varphi_{n-1}(s))ds

for

n \geq 1

step 2

To show that

\varphi_n(t)

are well-defined, we need to prove that the integral

\int_0^t (t-s)g(s,\varphi_{n-1}(s))ds

exists for

|t| \leq \alpha

step 3

Since

|g(t,y)| \leq M

and the functions are continuous, the integral is well-defined and finite for

|t| \leq \alpha

step 4

The value of

\alpha

is chosen to ensure that the integral and the sequence

\{\varphi_n(t)\}

remain bounded within the rectangle

R

step 5

To show convergence, we need to establish that the sequence

\{\varphi_n(t)\}

is Cauchy, which involves showing that the difference between successive terms goes to zero as

n

goes to infinity

step 6

If the sequence

\{\varphi_n(t)\}

converges, then it converges to a function

\varphi(t)

that satisfies the integral equation

y(t) = y_0 + z_0t + \int_0^t (t-s)g(s,y(s))ds

Answer

The sequence

\{\varphi_n(t)\}

is well-defined for

|t| \leq \alpha

and converges to a solution of the integral equation on

|t| \leq \alpha

.

Key Concept

Well-defined functions and convergence of function sequences

Explanation

The sequence

\{\varphi_n(t)\}

is constructed recursively with each term involving an integral that is well-defined due to the continuity and boundedness of

g(t,y)

. The convergence of this sequence to a function that satisfies the given integral equation is guaranteed by the properties of the functions involved and the choice of

\alpha

.

Consider the diﬀerential equation y0(t) = y2(t), y(0) = 1 on (−∞,1) (a) Show that on the region R = {(t,y)||t−0|≤ 2, |y−1|≤ b},b > 0, f(t,y) is bounded and ﬁnd the upper bound M.

Solution by Steps

step 1

To solve the differential equation

y'(t) = y(t)^2

with the initial condition

y(0) = 1

, we separate variables and integrate

step 2

Separating variables gives

\frac{dy}{y^2} = dt

step 3

Integrating both sides, we get

-\frac{1}{y} = t + C

step 4

Using the initial condition

y(0) = 1

, we find

C = -1

step 5

The solution to the differential equation is

y(t) = -\frac{1}{t-1}

Answer

y(t) = -\frac{1}{t-1}

Key Concept

Separation of Variables in Differential Equations

Explanation

To solve a first-order ordinary differential equation where the derivative of the function is equal to the function squared, we use separation of variables and integrate both sides. The initial condition is used to find the constant of integration.

Solution by Steps

step 1

To find the upper bound

M

of

f(t,y) = y^2

on the region

R = \{(t,y) \mid |t| \leq 2, |y-1| \leq b\}

, we evaluate the maximum value of

y

on

R

step 2

Since

|y-1| \leq b

, the maximum value of

y

is

1 + b

step 3

The upper bound

M

for

f(t,y)

is then

(1+b)^2

Answer

M = (1+b)^2

Key Concept

Finding the Upper Bound of a Function

Explanation

The upper bound of a function within a specified region can be found by evaluating the maximum value that the function can take within that region. In this case, the function

f(t,y) = y^2

is maximized when

y

is at its maximum value within the region

R

.

Consider the diﬀerential equation y0(t) = y2(t), y(0) = 1 on (−∞,1) (a) Show that on the region R = {(t,y)||t−0|≤ 2, |y−1|≤ b},b > 0, f(t,y) is bounded and ﬁnd the upper bound M. 1 (b) Given that α = mina, b M, ﬁnd the actual value of α. (c) Hence, determine the interval of existence of the solution to the diﬀerential equation. (d) Determine whether or not the diﬀerential equation satisfy the Lipschitz condition and if so, compute the Lipschitz constant.

Solution by Steps

step 1

To find the upper bound

M

for

f(t,y) = y^2

on the region

R = \{(t,y) \,|\, |t - 0| \leq 2, |y - 1| \leq b\}

, we need to consider the maximum value

y

can take in this region

step 2

Since

|y - 1| \leq b

, the maximum value of

y

is

1 + b

step 3

Substituting the maximum value of

y

into

f(t,y)

, we get

M = (1 + b)^2

Answer

M = (1 + b)^2

Key Concept

Finding the upper bound of a function on a closed region

Explanation

The upper bound is found by identifying the maximum value the function can take within the given constraints.

---

step 1

Given that

\alpha = \min(2, bM)

, we substitute the value of

M

from the previous solution

step 2

Substituting

M = (1 + b)^2

into the expression for

\alpha

, we get

\alpha = \min(2, b(1 + b)^2)

Answer

\alpha = \min(2, b(1 + b)^2)

Key Concept

Calculating the minimum of two expressions

Explanation

The actual value of

\alpha

is the smaller of the two values: 2 or

b(1 + b)^2

.

---

step 1

To determine the interval of existence of the solution to the differential equation

y'(t) = y^2

, we use the fact that the solution cannot continue past the vertical asymptote

step 2

The solution to the differential equation is

y(t) = \frac{1}{1 - t}

, which has a vertical asymptote at

t = 1

step 3

Since the initial condition is

y(0) = 1

and the solution is defined for t < 1 , the interval of existence is

(-\infty, 1)

Answer

The interval of existence is

(-\infty, 1)

.

Key Concept

Interval of existence for a differential equation

Explanation

The interval of existence is determined by the behavior of the solution and its vertical asymptotes.

---

step 1

To determine if

y'(t) = y^2

satisfies the Lipschitz condition on

R

, we need to check if the derivative of

f(t,y)

with respect to

y

is bounded on

R

step 2

The derivative of

f(t,y)

with respect to

y

is

2y

step 3

On the region

R

, the maximum value of

|2y|

is

2(1 + b)

, which is the Lipschitz constant

L

step 4

Since

L

is finite, the differential equation satisfies the Lipschitz condition on

R

Answer

The differential equation satisfies the Lipschitz condition with Lipschitz constant

L = 2(1 + b)

.

Key Concept

Lipschitz condition for a differential equation

Explanation

A differential equation satisfies the Lipschitz condition if the derivative of the function with respect to

y

is bounded on the given region.

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