Reread Example 3 and then discuss, with reference to Theorem...

Reread Example 3 and then discuss, with reference to Theorem $1.2 .1,$ the existence and uniqueness of a solution of the initial-value problem consisting of $x y^{\prime}-4 y=x^{6} e^{x}$ and the given initial condition. (a) $y(0)=0$ (b) $y(0)=y_{0}, y_{0} > 0$ (c) $y\left(x_{0}\right)=y_{0}, x_{0} > 0, y_{0} > 0$

Reread Example 3 and then discuss, with reference to Theorem $1.2 .1,$ the existence and uniqueness of a solution of the initial-value problem consisting of $x y^{\prime}-4 y=x^{6} e^{x}$ and the given initial condition.
(a) $y(0)=0$
(b) $y(0)=y_{0}, y_{0} > 0$
(c) $y\left(x_{0}\right)=y_{0}, x_{0} > 0, y_{0} > 0$

Key Concepts

Standard Form of a Linear Differential Equation

A first-order linear differential equation is typically written in the standard form y' + p(x)y = q(x). Expressing the equation in this form allows one to analyze the continuity of the coefficient functions and apply methods such as integrating factors. It also aids in determining whether the conditions for existence and uniqueness are met.

Singular Points in Differential Equations

A singular point is a point where one or more of the coefficient functions of the differential equation become discontinuous or are not defined (for example, due to division by zero). When an initial condition is specified at a singular point, as in some of the cases mentioned, the usual existence and uniqueness results may not apply directly, necessitating a more detailed analysis of the differential equation's behavior near that point.

Initial Value Problem

An initial value problem is a differential equation together with a condition specifying the value of the unknown function at a particular point. This initial condition is used to determine a unique solution, if one exists, among all possible solutions of the differential equation.

Existence and Uniqueness Theorem

The existence and uniqueness theorem, often associated with the Picard–Lindelöf theorem, provides conditions under which an initial value problem has a unique solution. Typically, if the function on the right-hand side of the differential equation and its partial derivative with respect to the dependent variable are continuous in a neighborhood of the initial condition, then the theorem guarantees a unique local solution.

Transcript

00:03 In this question we are given the initial value problem consisting of x5 prime minus 4y is equal to x raised x and exponential function x.

00:14 And the given conditions are a, that is, y of 0 is equal to 0, y of x not is equal to y0 for y0 greater than 0, and y0 of x .0 of x .0 is equal to y0 greater than 0 and y0 greater than 0.

00:30 So the first step is for a part, the first step is all the solution of the form, y is equal to x raised to power 5, exponential function minus x raised to par 4, exponential function, plus c x raised to power 4, satisfy the initial condition.

00:54 But 4 over x is discontinuous, is discontinuous.

01:04 At x is equal to 0 does the hypothesis of the theorem 1 .21 are not satisfied and the initial value problem does not have a unique solution so no unique solution now for part b the substitute x is equal to 0 and y is equal to y not in y is equal to n y is equal to n y y is equal to x raised to power 5 exponential function minus x x to power 4 exponential function plus c x x to power 4 and we obtain that y not is equal to 0 but it is given that y not is greater than 0 thus the differential equation has a no solution satisfying that y of 0 is equal to 0 0 that is y0 is greater than 0 so it means that this differential equation has no solution satisfying this condition it has no solution for it now the next thing is part c so the part c is for that we substitute x is equal to x0 and y is equal to y0 and y is equal to y0 and y is equal to x x x x x x x to par 4 exponential function plus c times x rd to par 4 and we obtain that y0 .0.

02:58 Is equal to x .0 5 exponential function minus x0 4 exponential function x0x0 plus c x04...

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