(x^2-9) dy/dx + 6y = (x+3)^2
A. Find the general solution of the given differential equation
B. Give the largest interval over which the general solution is defined
C. Determine whether there are any transient terms in the general solution
The correct answer and explanation is :
A. Finding the General Solution of the Differential Equation
The given differential equation is:
[
(x^2 – 9) \frac{dy}{dx} + 6y = (x+3)^2
]
This is a linear first-order differential equation, and it can be rewritten as:
[
\frac{dy}{dx} + \frac{6}{x^2 – 9} y = \frac{(x + 3)^2}{x^2 – 9}
]
This equation is in the form of:
[
\frac{dy}{dx} + P(x)y = Q(x)
]
where ( P(x) = \frac{6}{x^2 – 9} ) and ( Q(x) = \frac{(x + 3)^2}{x^2 – 9} ).
To solve this, we first compute the integrating factor ( \mu(x) ), which is given by:
[
\mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{6}{x^2 – 9} \, dx}
]
The denominator ( x^2 – 9 ) is a difference of squares, so we can simplify it as ( (x – 3)(x + 3) ). To integrate ( \frac{6}{x^2 – 9} ), we can use partial fraction decomposition:
[
\frac{6}{x^2 – 9} = \frac{A}{x – 3} + \frac{B}{x + 3}
]
By solving for ( A ) and ( B ), we get:
[
A = 1, \quad B = -1
]
Thus, the integral becomes:
[
\int \frac{6}{x^2 – 9} \, dx = \int \left( \frac{1}{x – 3} – \frac{1}{x + 3} \right) \, dx
]
This gives:
[
\mu(x) = e^{\ln|x – 3| – \ln|x + 3|} = \frac{|x – 3|}{|x + 3|}
]
Now, multiply both sides of the original equation by the integrating factor:
[
\frac{|x – 3|}{|x + 3|} \frac{dy}{dx} + \frac{6}{x^2 – 9} \cdot \frac{|x – 3|}{|x + 3|} y = \frac{(x + 3)^2}{x^2 – 9} \cdot \frac{|x – 3|}{|x + 3|}
]
After solving this equation, the general solution will be:
[
y(x) = C \frac{|x + 3|}{|x – 3|}
]
where ( C ) is a constant of integration.
B. Largest Interval over Which the General Solution is Defined
The solution contains terms involving ( |x – 3| ) and ( |x + 3| ), and these terms are undefined at ( x = 3 ) and ( x = -3 ). Thus, the solution is defined on the largest interval where both ( x = 3 ) and ( x = -3 ) are not included. Therefore, the largest interval over which the general solution is defined is:
[
(-\infty, -3) \cup (3, \infty)
]
C. Transient Terms in the General Solution
Transient terms refer to terms in the solution that decay over time or disappear as ( x ) approaches infinity or negative infinity. In this case, the solution does not contain exponential terms or terms that decay to zero as ( x ) tends to infinity. The general solution contains terms that depend on ( x ) in an algebraic form (i.e., ( \frac{|x + 3|}{|x – 3|} )) and does not have any transient behavior that would disappear with increasing or decreasing ( x ).
Therefore, there are no transient terms in the general solution. The solution is purely algebraic and will not decay as ( x ) increases or decreases.