Find The Solution To The Initial-value Problem 3xy' - 5y = 7x^3; Y(1) = C. ; Y = is a common problem encountered in differential equations, particularly in the study of first-order linear differential equations. In this article, we will explore step-by-step how to solve this initial-value problem (IVP), understand the underlying methods, and interpret the solution. Whether you're a student preparing for exams or a professional applying differential equations in modeling, understanding this problem will enhance your problem-solving toolkit.
Understanding the Initial-value Problem
Before diving into the solution, it's essential to understand what the problem entails. The differential equation provided is:
3xy' - 5y = 7x^3
with the initial condition:
Y(1) = C
and the goal is to find the explicit form of the function \(Y = y(x)\) satisfying both the differential equation and the initial condition.
Rearranging and Recognizing the Type of Differential Equation
Standard Form of a First-Order Linear Differential Equation
A first-order linear differential equation has the general form:
y' + P(x) y = Q(x)
To analyze our problem, we need to rewrite the given differential equation into this standard form.
Given:
3xy' - 5y = 7x^3
Divide both sides by 3x (assuming \(x \neq 0\)):
y' - \frac{5}{3x} y = \frac{7x^2}{3}
Now, the equation is in the form:
y' + P(x) y = Q(x)
where:
- \(P(x) = -\frac{5}{3x}\)
- \(Q(x) = \frac{7x^2}{3}\)
Type Identification
This is a linear first-order differential equation that can be solved using an integrating factor method.
Methodology for Solving the Differential Equation
Step 1: Find the Integrating Factor (IF)
The integrating factor is given by:
\(\mu(x) = e^{\int P(x) dx}\)
Compute:
\[
\int P(x) dx = \int -\frac{5}{3x} dx = -\frac{5}{3} \int \frac{1}{x} dx = -\frac{5}{3} \ln|x|
\]
Thus:
\[
\mu(x) = e^{-\frac{5}{3} \ln|x|} = |x|^{-\frac{5}{3}}
\]
Since we're considering \(x > 0\) (from initial condition \(x=1\)), we can write:
\[
\mu(x) = x^{-\frac{5}{3}}
\]
Step 2: Multiply the entire differential equation by the integrating factor
Multiplying:
\[
x^{-\frac{5}{3}} y' - \frac{5}{3x} x^{-\frac{5}{3}} y = \frac{7x^2}{3} x^{-\frac{5}{3}}
\]
which simplifies to:
\[
\frac{d}{dx} \left( x^{-\frac{5}{3}} y \right) = \frac{7x^2}{3} x^{-\frac{5}{3}}
\]
because the left side is the derivative of \( \mu(x) y \).
Step 3: Integrate both sides to find \( y(x) \)
Integrate:
\[
x^{-\frac{5}{3}} y = \int \frac{7x^2}{3} x^{-\frac{5}{3}} dx + K
\]
Express the integrand:
\[
\frac{7}{3} x^{2} \cdot x^{-\frac{5}{3}} = \frac{7}{3} x^{2 - \frac{5}{3}} = \frac{7}{3} x^{\frac{6}{3} - \frac{5}{3}} = \frac{7}{3} x^{\frac{1}{3}}
\]
Now, the integral becomes:
\[
\int \frac{7}{3} x^{\frac{1}{3}} dx = \frac{7}{3} \int x^{\frac{1}{3}} dx
\]
Recall:
\[
\int x^{n} dx = \frac{x^{n+1}}{n+1} + C
\]
So:
\[
\frac{7}{3} \cdot \frac{x^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} = \frac{7}{3} \cdot \frac{x^{\frac{4}{3}}}{\frac{4}{3}} = \frac{7}{3} \times \frac{3}{4} x^{\frac{4}{3}} = \frac{7}{4} x^{\frac{4}{3}}
\]
Therefore,
\[
x^{-\frac{5}{3}} y = \frac{7}{4} x^{\frac{4}{3}} + K
\]
where \(K\) is the constant of integration.
Expressing the General Solution
Multiply both sides by \( x^{\frac{5}{3}} \):
\[
y(x) = x^{\frac{5}{3}} \left( \frac{7}{4} x^{\frac{4}{3}} + K \right) = \frac{7}{4} x^{\frac{5}{3} + \frac{4}{3}} + K x^{\frac{5}{3}}
\]
Simplify exponents:
\[
\frac{5}{3} + \frac{4}{3} = \frac{9}{3} = 3
\]
Thus,
\[
\boxed{
Y(x) = y(x) = \frac{7}{4} x^{3} + C x^{\frac{5}{3}}
}
\]
where \( C \) is an arbitrary constant (replacing \(K\)).
Applying the Initial Condition
Given:
\[
Y(1) = C
\]
Substitute \( x=1 \):
\[
Y(1) = \frac{7}{4} \times 1^{3} + C \times 1^{\frac{5}{3}} = \frac{7}{4} + C
\]
Since the initial condition specifies \( Y(1) = C \), we have:
\[
C = \frac{7}{4} + C
\]
This implies:
\[
C = \frac{7}{4} + C
\]
which leads to:
\[
0 = \frac{7}{4}
\]
This indicates a contradiction unless the notation is clarified. To avoid confusion, let's denote the arbitrary constant in the general solution as \(k\), and the constant in the initial condition as \(C\):
- General solution:
\[
Y(x) = \frac{7}{4} x^{3} + k x^{\frac{5}{3}}
\]
- Initial condition:
\[
Y(1) = C
\]
Substitute:
\[
C = \frac{7}{4} \times 1^{3} + k \times 1^{\frac{5}{3}} = \frac{7}{4} + k
\]
Solve for \(k\):
\[
k = C - \frac{7}{4}
\]
Final solution incorporating initial condition:
\[
\boxed{
Y(x) = \frac{7}{4} x^{3} + \left(C - \frac{7}{4}\right) x^{\frac{5}{3}}
}
\]
This represents the explicit solution to the initial-value problem.
Summary of the Solution Process
- Rearranged the differential equation into standard linear form.
- Calculated the integrating factor \( \mu(x) = x^{-\frac{5}{3}} \).
- Multiplied through by the integrating factor to facilitate integration.
- Integrated both sides to find the general solution.
- Applied the initial condition to determine the specific solution constant.
Additional Insights and Applications
Why Is This Solution Important?
Understanding how to solve this IVP provides foundational skills applicable in various fields, including physics, engineering, and economics. Differential equations describe systems where change depends on current states, such as population dynamics, heat transfer, and electrical circuits.
Potential Variations and Generalizations
- Different initial conditions: Altering the value of \(Y(1)\) changes the particular solution
