Determine the longest interval in which the given initial value

smismSitlougsyy

smismSitlougsyy

Answered question

2021-11-11

Determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution.
(x+3)y″+xy′+(ln|x|)y=0, y(1)=0, y′(1)=1

Answer & Explanation

Mike Henson

Mike Henson

Beginner2021-11-12Added 11 answers

(x+3)y ″+xy′+(ln|x|)y=0
y ″+xx+3y′+ln|x|x+3y=0
y ″+p(x)y′+q(x)y=g(x)
y(1)=0,y′(1)=1
We use notation of Theorem,
1, for p(x)=xx+3, it is continuous on (−∞,−3) and (−3,∞)
2. for q(x)=ln|x|x+3, it is continuous on (−∞,−3),(−3,0) and (0,∞).
3. for g(x)=0, it is continuous on (−∞,∞)
We have t0=1, then the longest interval of the solution is (0,∞)

karton

karton

Expert2023-05-09Added 613 answers

To determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution, we will solve the differential equation without finding the solution explicitly. The initial value problem is defined as:
(x+3)y″+xy′+(ln|x|)y=0,y(1)=0,y′(1)=1
To proceed, we can rewrite the differential equation in a more standard form by dividing through by (x+3):
y″+xx+3y′+ln|x|x+3y=0
Now, let's consider the interval in which the given initial condition is defined, which is x>0 since the natural logarithm ln|x| is only defined for positive values of x.
Next, we'll analyze the coefficient functions involved in the differential equation. The coefficient xx+3 is continuous and bounded on the interval x>0. Additionally, the coefficient ln|x|x+3 is also continuous and bounded on the interval x>0.
Now, let's examine the initial conditions. We have y(1)=0, which means the solution must pass through the point (1,0). Additionally, we have y′(1)=1, which implies that the slope of the solution curve at x=1 is 1.
Based on the continuity and boundedness of the coefficient functions, along with the given initial conditions, we can conclude that a unique twice differentiable solution exists on the interval 0<x<a, where a is the largest positive number for which the coefficients remain continuous and bounded.
Therefore, the longest interval in which the initial value problem is certain to have a unique twice differentiable solution is 0<x<a.
user_27qwe

user_27qwe

Skilled2023-05-09Added 375 answers

The existence and uniqueness theorem for second-order linear differential equations can be used to calculate the longest interval in which the given starting value issue is guaranteed to have a unique twice differentiable solution. According to the theorem, a second-order linear homogeneous differential equation has the following form:
a(x)y″+b(x)y′+c(x)y=0,
where a(x), b(x), and c(x) are continuous functions on an interval I, and if the initial conditions y(x0)=y0 and y′(x0)=y0′ are specified at some point x0 within I, then there exists a unique solution y(x) that is twice differentiable on I.
In the given initial value problem, we have:
(x+3)y″+xy′+(ln|x|)y=0,
with initial conditions y(1)=0 and y′(1)=1. To apply the existence and uniqueness theorem, we need to check the continuity of the coefficients a(x)=(x+3), b(x)=x, and c(x)=ln|x| on an interval I.
First, let's consider the coefficient a(x)=(x+3). This is a polynomial function and is continuous for all x.
Next, we have the coefficient b(x)=x, which is also a polynomial function and is continuous for all x.
Finally, we have the coefficient c(x)=ln|x|. This function is not defined at x=0, so we need to make sure our interval I does not include x=0. Additionally, we need to consider the continuity of ln|x| for x≠0. The natural logarithm function ln(x) is continuous for positive x, and for negative x, we have ln(−x)=ln|x|+iπ, where i is the imaginary unit. Therefore, ln|x| is not continuous at x=0 but is continuous for x≠0.
Now, let's determine the longest interval for which the given initial value problem has a unique twice differentiable solution. Since the coefficients a(x), b(x), and c(x) are all continuous on (−∞,0)∪(0,∞) (excluding x=0), we can choose an interval I within this range.
Thus, the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution is I=(−∞,0)∪(0,∞).
Please note that this solution assumes x≠0 due to the discontinuity of ln|x| at x=0.
Don Sumner

Don Sumner

Skilled2023-05-09Added 184 answers

The given initial value problem is:
(x+3)y″+xy′+(ln|x|)y=0,y(1)=0,y′(1)=1.
To solve this second-order linear homogeneous differential equation, we can use the method of Frobenius. Let's assume a power series solution of the form:
y(x)=∑n=0∞an(x−1)n+r.
Substituting this into the differential equation, we can find an expression for the coefficients an. Differentiating y(x), we have:
y′(x)=∑n=0∞an(n+r)(x−1)n+r−1.
Similarly, differentiating again, we get:
y″(x)=∑n=0∞an(n+r)(n+r−1)(x−1)n+r−2.
Now, let's substitute these expressions into the differential equation and simplify:
(x+3)(∑n=0∞an(n+r)(n+r−1)(x−1)n+r−2)+x(∑n=0∞an(n+r)(x−1)n+r−1)+(ln|x|)(∑n=0∞an(x−1)n+r)=0.
Next, we expand the terms and combine like powers of (x−1):
∑n=0∞an(n+r)(n+r−1)(x−1)n+r+3∑n=0∞an(n+r)(n+r−1)(x−1)n+r−2+∑n=0∞an(n+r)(x−1)n+r+ln|x|∑n=0∞an(x−1)n+r=0.
We can now combine the terms and rewrite the equation as:
∑n=0∞[an(n+r)(n+r−1)(x−1)n+r+3an(n+r)(n+r−1)(x−1)n+r−2+an(n+r)(x−1)n+r]+ln|x|∑n=0∞an(x−1)n+r=0.
To ensure the above equation holds for all x, each term within the series must vanish independently. This leads to the following recurrence relation for the coefficients an:
an(n+r)(n+r−1)+3an(n+r)(n+r−1)+an(n+r)=0.
Simplifying the equation, we have:
an[(n+r)(n+r−1)+3(n+r)(n+r−1)+(n+r)]=0.
Since we are looking for a non-trivial solution, the coefficient an cannot be zero.
Therefore, the expression in the brackets must equal zero:
(n+r)(n+r−1)+3(n+r)(n+r−1)+(n+r)=0.
We can simplify this equation as:
[(n+r)+3(n+r)+1](n+r−1)=0.
Simplifying further, we have:
(n+r)(4n+4r)=0.
This equation gives us two possibilities:
1. n+r=0
2. 4n+4r=0
For case 1, when n+r=0, we obtain n=−r. This means that the indices of the series are given by n=−r, n=−r+1, n=−r+2, and so on.
For case 2, when 4n+4r=0, we find n=−r. However, this case is equivalent to the one we found in case 1.
Therefore, the indices of the series are given by n=−r, n=−r+1, n=−r+2, and so on.
Now, let's consider the initial conditions y(1)=0 and y′(1)=1. Substituting x=1 into the power series solution, we have:
y(1)=∑n=0∞an(1−1)n+r=a0·0n+r=a0·0=0.
Therefore, we can conclude that a0=0.
Differentiating the power series solution and substituting x=1, we get:
y′(1)=∑n=0∞an(n+r)(1−1)n+r−1=∑n=0∞an(n+r)(0)n+r−1=0.
From this equation, we can conclude that all the terms in the series involving an vanish. However, we know that a0=0. This implies that all the coefficients an for n≥1 must also be zero.
Therefore, the power series solution reduces to y(x)=a0(x−1)−r.
Since we have a0=0, the power series solution is identically zero. Hence, the only solution to the initial value problem is the trivial solution y(x)=0.
In summary, the given initial value problem does not have a non-trivial solution, and therefore, there is no interval in which a unique twice differentiable solution exists.
Vasquez

Vasquez

Expert2023-05-14Added 669 answers

Result:
x∈(−∞,−3)∪(−3,0)∪(0,∞).
Solution:
To determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution, we need to analyze the coefficients and the singularities of the equation.
The given initial value problem is:
(x+3)y′+xy+(ln|x|)′y=0,y(1)=0,y′(1)=0.
Let's examine the coefficients of the equation:
- The coefficient of y′ is (x+3).
- The coefficient of y is xy.
- The coefficient of y multiplied by the derivative of the natural logarithm of |x| is (ln|x|)′y.
We observe that the coefficient of y′, which is (x+3), is a polynomial function that is defined for all values of x except x=−3. Therefore, the coefficient of y′ is well-defined for all x≠−3.
Next, we consider the coefficient of y, which is xy. This coefficient is a polynomial function defined for all values of x. Hence, the coefficient of y is well-defined for all x.
Finally, we analyze the coefficient of y multiplied by the derivative of the natural logarithm of |x|, which is (ln|x|)′y. The derivative of the natural logarithm of |x| is not defined at x=0 since the absolute value function has a singularity there. Consequently, the coefficient of y multiplied by (ln|x|)′ is not well-defined for x=0.
In summary, the coefficient of y′ is well-defined for all x≠−3, the coefficient of y is well-defined for all x, and the coefficient of y multiplied by (ln|x|)′ is not well-defined for x=0. Therefore, the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution is:
x∈(−∞,−3)∪(−3,0)∪(0,∞).

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