Problem 8
Question
Classify each differential equation as separable, exact, linear, homogeneous, or Bernoulli. Some equations may be more than one kind. Do not solve. (a) \(\frac{d y}{d x}=\frac{x-y}{x}\) (b) \(\frac{d y}{d x}=\frac{1}{y-x}\) (c) \((x+1) \frac{d y}{d x}=-y+10\) (d) \(\frac{d y}{d x}=\frac{1}{x(x-y)}\) (e) \(\frac{d y}{d x}=\frac{y^{2}+y}{x^{2}+x}\) (f) \(\frac{d y}{d x}=5 y+y^{2}\) (g) \(y d x=\left(y-x y^{2}\right) d y\) (h) \(x \frac{d y}{d x}=y e^{x y}-x\) (i) \(x y y^{\prime}+y^{2}=2 x\) (j) \(2 x y y^{\prime}+y^{2}=2 x^{2}\) (k) \(y d x+x d y=0\) (l) \(\left(x^{2}+\frac{2 y}{x}\right) d x=\left(3-\ln x^{2}\right) d y\) (m) \(\frac{d y}{d x}=\frac{x}{y}+\frac{y}{x}+1\) (n) \(\frac{y}{x^{2}} \frac{d y}{d x}+e^{2 x^{3}+y^{2}}=0\)
Step-by-Step Solution
Verified Answer
Classifications: (a) Linear, (b) None, (c) Linear, (d) None, (e) Separable, (f) Bernoulli, (g) Separable, (h) None, (i) None, (j) None, (k) Exact, (l) Homogeneous, (m) Separable, (n) None.
1Step 1: Analyze Equation (a)
The equation \( \frac{d y}{d x}=\frac{x-y}{x} \) can be rewritten in the form \( \frac{d y}{d x} + \frac{1}{x}y = 1 \), which is linear in the standard form \( \frac{d y}{d x} + P(x)y = Q(x) \). It is not separable, exact, homogeneous, or Bernoulli.
2Step 2: Analyze Equation (b)
The equation \( \frac{d y}{d x}=\frac{1}{y-x} \) cannot be directly separated or considered exact. It is not linear or homogeneous because of the lack of clear structure fitting those types, nor is it Bernoulli. It's not classifiable into standard categories without further transformation.
3Step 3: Analyze Equation (c)
The equation \( (x+1) \frac{d y}{d x}=-y+10 \) can be rearranged to \( \frac{d y}{d x} + \frac{1}{x+1}y = \frac{10}{x+1} \), which is linear in the standard form \( \frac{d y}{d x} + P(x)y = Q(x) \). It's not separable, exact, homogeneous, or Bernoulli.
4Step 4: Analyze Equation (d)
The equation \( \frac{d y}{d x}=\frac{1}{x(x-y)} \) cannot be separated easily nor described as exact, linear, or a Bernoulli type with its current structure.
5Step 5: Analyze Equation (e)
The equation \( \frac{d y}{d x}=\frac{y^{2}+y}{x^{2}+x} \) can be separated into terms involving only \( x \) and \( y \), making it a separable differential equation.
6Step 6: Analyze Equation (f)
The equation \( \frac{d y}{d x}=5 y+y^{2} \) fits the form of a Bernoulli equation, \( \frac{d y}{d x} + P(x)y = Q(x) y^n \), with \( n = 2 \), a special case of nonlinear equations. It can also be considered separable.
7Step 7: Analyze Equation (g)
The equation \( y d x = (y-x y^{2}) d y \) is separable as it can be transformed into \( \frac{dx}{y} = \frac{dy}{y-x y^2} \), involving only terms in \( x \) or \( y \).
8Step 8: Analyze Equation (h)
The equation \( x \frac{d y}{d x} = y e^{x y} - x \), with its complex exponential term depending on \( x \) and \( y \), does not fit into separable, linear, or other mentioned categories easily.
9Step 9: Analyze Equation (i)
The equation \( x y y^{\prime}+y^{2}=2 x \) can't be directly classified due to non-linear and non-standard structure. It may be manipulated into a type but does not fit easy categories here.
10Step 10: Analyze Equation (j)
In \( 2 x y y^{\prime}+y^{2}=2 x^{2} \), the structure does not permit straightforward classification into the standard types due to multiplicative terms involving \( y \) and its derivative.
11Step 11: Analyze Equation (k)
The equation \( y dx + x dy = 0 \) is an exact equation because \( \partial N/\partial x = \partial M/\partial y = 1\), satisfying the condition for an exact differential.
12Step 12: Analyze Equation (l)
The equation \( \left(x^{2}+\frac{2 y}{x}\right) d x=\left(3-\ln x^{2}\right) d y \), after manipulation, fits the structure of a homogeneous equation where variables can be grouped into the form \( d(y/x) \).
13Step 13: Analyze Equation (m)
The equation \( \frac{d y}{d x}=\frac{x}{y}+\frac{y}{x}+1 \) can be rearranged into a form that can be treated as separable or Bernoulli upon additional rearrangements, inheriting characteristics of both.
14Step 14: Analyze Equation (n)
The equation \( \frac{y}{x^{2}} \frac{d y}{d x} + e^{2 x^{3}+y^{2}} = 0 \), doesn't fit directly into these typical categories due to its exponential term depending nontrivially on \( x \) and \( y \).
Key Concepts
Separable Differential EquationsExact Differential EquationsLinear Differential EquationsBernoulli Differential EquationsHomogeneous Differential Equations
Separable Differential Equations
Separable differential equations are a category of equations where the variables can be separated on each side of the equation, with one side in terms of one variable and the other in terms of its counterpart. The main idea is to rewrite the equation into the form where the derivative becomes a product of a function of the independent variable and a function of the dependent variable. This allows us to integrate both sides separately.
For example, equation (e) from the example \[ \frac{d y}{d x} = \frac{y^2 + y}{x^2 + x} \]can be rewritten as:\[ \frac{dy}{y^2 + y} = \frac{dx}{x^2 + x} \]This form illustrates how these two sides can be integrated separately. Equations that separate in such a way can be solved by integrating both sides, leading eventually to solutions that depend on integration constants.
For example, equation (e) from the example \[ \frac{d y}{d x} = \frac{y^2 + y}{x^2 + x} \]can be rewritten as:\[ \frac{dy}{y^2 + y} = \frac{dx}{x^2 + x} \]This form illustrates how these two sides can be integrated separately. Equations that separate in such a way can be solved by integrating both sides, leading eventually to solutions that depend on integration constants.
Exact Differential Equations
An exact differential equation involves two functions and their derivatives that satisfy a specific condition. These equations typically have the form:\[ M(x, y)\,dx + N(x, y)\,dy = 0 \]For an equation to be exact, the partial derivative of \(M\) with respect to \(y\) must equal the partial derivative of \(N\) with respect to \(x\) (\(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)). This ensures that there is some potential function \(\phi(x, y)\) such that \(d\phi(x, y) = 0\), meaning the equation represents a total differential.
An example from the exercise is equation (k):\[ y\,dx + x\,dy = 0 \]Here, \(M = y\) and \(N = x\), and since both are constant values, the condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) is inherently satisfied, confirming its exactness.
An example from the exercise is equation (k):\[ y\,dx + x\,dy = 0 \]Here, \(M = y\) and \(N = x\), and since both are constant values, the condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) is inherently satisfied, confirming its exactness.
Linear Differential Equations
Linear differential equations are those where the dependent variable and its derivative appear to the first power only, without products or nonlinear functions of the dependent variable. They can be generally expressed as:\[ \frac{dy}{dx} + P(x)y = Q(x) \]This straightforward linear structure implies methods like integrating factors can be applied to solve them. Looking at equation (a) from the problem:\[ \frac{dy}{dx} + \frac{1}{x}y = 1 \]This is linear, where \(P(x) = \frac{1}{x}\) and \(Q(x) = 1\). Linear equations are especially important because of their predictability and broad applicability in scientific fields.
Bernoulli Differential Equations
Bernoulli differential equations feature a unique structure that introduces nonlinear elements, requiring a transformation for solution. They adhere to the form:\[ \frac{dy}{dx} + P(x)y = Q(x)y^n \]with \(n\) not equal to 0 or 1, making it a nonlinear equation. This distinct form allows for a substitution, typically \(z = y^{1-n}\), transforming the equation into a linear one.
An example from our exercise would be equation (f):\[ \frac{dy}{dx} = 5y + y^2 \]which can be reformulated as a Bernoulli equation:\[ \frac{dy}{dx} + (-5)y = y^2 \]This can be solved using a substitution, changing it into a standard linear form and solving accordingly.
An example from our exercise would be equation (f):\[ \frac{dy}{dx} = 5y + y^2 \]which can be reformulated as a Bernoulli equation:\[ \frac{dy}{dx} + (-5)y = y^2 \]This can be solved using a substitution, changing it into a standard linear form and solving accordingly.
Homogeneous Differential Equations
Homogeneous differential equations involve terms that can be expressed as a ratio, typically achieved by substituting variables into new forms that exploit the ratio. In essence, both sides of the differential equation can be rewritten in terms of new coordinates or parameters.The typical form is:\[ \frac{dy}{dx} = f\left(\frac{y}{x}\right) \]Equation (l) from our example showcases a structure modifiable into a homogeneous form:\[ (x^2 + \frac{2y}{x})dx = (3 - \ln{x^2})dy \]With proper substitution, the differential equation can be expressed in terms of the ratio \(\frac{y}{x}\), facilitating a solution through integration. Homogeneous equations lend themselves well to a specific category of substitutions that simplify these complexities into integrable forms.
Other exercises in this chapter
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