In Each Of Problems 4 Through 9, Find The General Solution Of The Given Differential Equation. In Problems
When tackling differential equations, a fundamental step is to determine their general solutions, which encompass all possible solutions to the given problem. These solutions often depend on arbitrary constants, reflecting the infinite set of functions satisfying the differential equation. Problems 4 through 9 typically involve various types of differential equations, including separable, linear, exact, and homogeneous equations. Understanding the methods to solve each type is crucial for systematically deriving the general solution. This article explores a step-by-step approach to solving these equations, emphasizing key techniques and common pitfalls.
Problem 4: Solving a Separable Differential Equation
Understanding Separable Differential Equations
A differential equation is called separable if it can be written in the form:
\[
\frac{dy}{dx} = f(x)g(y)
\]
which allows the variables to be separated on either side:
\[
\frac{1}{g(y)} dy = f(x) dx
\]
The main goal is to integrate both sides to find the general solution.
Methodology for Solving
- Rewrite the differential equation in the separable form.
- Separate the variables: all y's on one side, all x's on the other.
- Integrate both sides: \(\int \frac{1}{g(y)} dy = \int f(x) dx + C\).
- Solve the resulting equation for y explicitly if possible, or leave it in implicit form.
Example
Suppose the differential equation is:
\[
\frac{dy}{dx} = y^2 \sin x
\]
Solution:
- Rewrite as:
\[
\frac{1}{y^2} dy = \sin x dx
\]
- Integrate both sides:
\[
\int y^{-2} dy = \int \sin x dx
\]
which yields:
\[
- y^{-1} = - \cos x + C
- Rearranged:
\[
\frac{1}{y} = \cos x + C'
\]
where \(C' = -C\). The general solution:
\[
y(x) = \frac{1}{\cos x + C'}
\]
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Problem 5: Solving a Linear Differential Equation
Understanding Linear Differential Equations
A first-order linear differential equation has the form:
\[
\frac{dy}{dx} + P(x) y = Q(x)
\]
where \(P(x)\) and \(Q(x)\) are continuous functions on an interval.
Methodology for Solution
- Identify \(P(x)\) and \(Q(x)\).
- Find the integrating factor (IF):
\[
\mu(x) = e^{\int P(x) dx}
\] - Multiply the entire differential equation by \(\mu(x)\), transforming it into:
\[
\frac{d}{dx} \left( \mu(x) y \right) = \mu(x) Q(x)
\] - Integrate both sides:
\[
\mu(x) y = \int \mu(x) Q(x) dx + C
\] - Finally, solve for \(y\):
\[
y = \frac{1}{\mu(x)} \left( \int \mu(x) Q(x) dx + C \right)
\]
Example
Given:
\[
\frac{dy}{dx} + 2 y = e^{x}
\]
- \(P(x) = 2\), \(Q(x) = e^{x}\).
- Integrating factor:
\[
\mu(x) = e^{\int 2 dx} = e^{2x}
\]
- Multiply the differential equation:
\[
e^{2x} \frac{dy}{dx} + 2 e^{2x} y = e^{3x}
\]
or
\[
\frac{d}{dx} \left( e^{2x} y \right) = e^{3x}
\]
- Integrate:
\[
e^{2x} y = \int e^{3x} dx + C = \frac{1}{3} e^{3x} + C
\]
- Solve for \(y\):
\[
y = e^{-2x} \left( \frac{1}{3} e^{3x} + C \right) = \frac{1}{3} e^{x} + C e^{-2x}
\]
This is the general solution.
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Problem 6: Solving an Exact Differential Equation
Understanding Exact Equations
An equation of the form:
\[
M(x, y) dx + N(x, y) dy = 0
\]
is exact if:
\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
\]
In such cases, there exists a potential function \(\Psi(x, y)\) such that:
\[
\frac{\partial \Psi}{\partial x} = M(x, y), \quad \frac{\partial \Psi}{\partial y} = N(x, y)
\]
and the general solution is:
\[
\Psi(x, y) = C
\]
Methodology for Solving
- Verify whether the differential equation is exact: check the equality of mixed partial derivatives.
- If exact, find \(\Psi(x, y)\) by integrating \(M(x, y)\) with respect to \(x\), treating \(y\) as constant.
- Differentiate \(\Psi\) with respect to \(y\) and set equal to \(N(x, y)\) to find any functions of \(x\) or \(y\) involved.
- Write the implicit solution as \(\Psi(x, y) = C\).
Example
Given:
\[
(2xy + y^2) dx + (x^2 + 2xy) dy = 0
\]
- \(M(x, y) = 2xy + y^2\), \(N(x, y) = x^2 + 2xy\).
- Check exactness:
\[
\frac{\partial M}{\partial y} = 2x + 2y
\]
\[
\frac{\partial N}{\partial x} = 2x + 2y
\]
They are equal, so the equation is exact.
- Find \(\Psi(x, y)\):
\[
\Psi_x = M = 2xy + y^2
\]
Integrate with respect to \(x\):
\[
\Psi(x, y) = \int (2xy + y^2) dx = y \int 2x dx + y^2 \int dx = y x^2 + y^2 x + h(y)
\]
- Differentiate \(\Psi\) with respect to \(y\):
\[
\Psi_y = x^2 + 2 y x + h'(y)
\]
Set equal to \(N(x, y) = x^2 + 2 xy\):
\[
x^2 + 2 y x + h'(y) = x^2 + 2 xy
\]
which simplifies to:
\[
h'(y) = 0
\]
Thus, \(h(y) = \text{constant}\).
- The implicit solution:
\[
\Psi(x, y) = y x^2 + y^2 x = C
\]
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Problem 7: Homogeneous Differential Equations
Understanding Homogeneous Equations
A first-order differential equation is homogeneous if it can be expressed as:
\[
\frac{dy}{dx} = F\left( \frac{y}{x} \right)
\]
or equivalently, the right-hand side is a homogeneous function of degree zero.
Methodology for Solution
- Make the substitution \(v = y/x\), which implies \(y = v x\).
- Differentiate \(y = v x\):
\[
\frac{dy}{dx} = v + x \frac{dv}{dx}
\] - Substitute into the original equation:
\[
v + x \frac{dv}{dx} = F(v)
\] - Solve the resulting separable differential equation:
\[
x \frac{dv}{dx} = F(v) - v
\]which can be written as:
\[
\frac{dv}{F(v) - v} = \frac{dx}{x}
\] - Integrate both sides to find \(v\) in terms of \(x\), then back-substitute \(
