Differential Equations · Lesson 6 of 6

Chapter Summary and Practice

“Connect classification, solution families, and all three solving methods through mixed reasoning and practice.”

Learning Objectives

• Connect order, degree, solution families, and initial conditions across the chapter. • Choose and justify a suitable method for an unfamiliar first-order equation. • Use separation, ratio substitutions, and integrating factors in mixed problems. • Verify implicit and explicit results and check domains and excluded solutions. • Translate geometric and growth statements into rate equations. • Use substitutions and historical context to understand the relationships between the methods.

See the whole chain of reasoning

The chapter began with a question about change: which function has a prescribed derivative relationship? We can now give that question a complete workflow. Read the equation, decide what sort of information it contains, choose a method that matches its structure, and check the resulting function rather than trusting the integration alone.

Order and degree describe an equation; they do not automatically tell you how to solve it. A solution is a function on an appropriate interval. A general family contains arbitrary constants, while a particular solution uses conditions to fix them. The three main first-order methods are linked because each ultimately turns the differential equation into an integral.

Idea	What to look for	Connection to the next step
Differential equation	A derivative of a dependent variable	The unknown is a function
Order	Highest derivative present	Standard general families have corresponding independent constants
Degree	Polynomial dependence on all derivatives	Distinct from order and from linearity
Verification	Differentiate and substitute	Checks a candidate without needing a solving method
General and particular solutions	Arbitrary constants and given conditions	A condition selects a family member
Separable	$y^{'} = g (x) h (y)$	Integrate one-variable expressions
Homogeneous	$y^{'} = g (y / x)$	A ratio substitution makes the equation separable
Linear	$y^{'} + P (x) y = Q (x)$	A multiplier makes a product derivative
Reverse variable	$d x / d y$ may be simpler	Use the same methods with y independent
Tangent geometry	Slope equals $d y / d x$	Translate the description before solving
Proportional growth	$P^{'} = k P$	Separation gives an exponential
Final check	Original equation, point, interval, excluded factors	Confirms what the result actually solves

For separation, gather the $y$ -dependent factor with $d y$ and the $x$ -dependent factor with $d x$ . For homogeneity, replace the ratio with a variable and use the product rule. For linear equations, normalise and use $e^{\int P d x}$ . These routes can overlap: $y^{'} = x (1 + y)$ is both separable and linear, so either correct method gives the same family.

Three integration routesLaTeX

Separable: Homogeneous: Linear: \int \frac{d y}{h ( y )} = \int g (x) d x + C y = v x, x \frac{d v}{d x} = g (v) - v μ = e^{\int P d x}, μ y = \int μ Q d x + C

Use each formula only after checking its assumptions. For the reverse-variable forms, interchange the roles of x and y consistently.

Classify and verify before solving

A mixed set may ask only for order and degree or for verification of a supplied family. Do the task requested. A long-looking formula may be easy to verify, and an equation of undefined degree may still be solvable after a useful rearrangement.

Example — Separate order from degree again

Problem
Classify $y^{''} + 5 x (y^{'})^{2} - 6 y = ln x$ , $(y^{'})^{3} - 4 (y^{'})^{2} + 7 y = sin x$ , and $y^{(4)} - sin (y^{'''}) = 0$ .

1.The first has highest derivative $y^{''}$ and is polynomial in its derivatives: order $2$ , degree $1$ . Its displayed logarithm also requires $x > 0$ .
2.The second contains only the first derivative, raised to a maximum power of $3$ : order $1$ , degree $3$ .
3.The third contains the fourth derivative, so order is $4$ . The sine of a derivative makes degree undefined.

Example — Verify an implicit family

Problem
Verify that $x^{2} - y^{2} = c (x^{2} + y^{2})^{2}$ satisfies $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ away from the origin.

1.Differentiate the proposed family, keeping c constant: $2 x d x - 2 y d y = 4 c (x^{2} + y^{2}) (x d x + y d y)$ .
2.Use $c = (x^{2} - y^{2}) / (x^{2} + y^{2})^{2}$ , valid away from the origin, and divide the differentiated equation by $2$ .
3.After multiplication by $x^{2} + y^{2}$ , the relation is $(x d x - y d y) (x^{2} + y^{2}) = 2 (x^{2} - y^{2}) (x d x + y d y)$ .
4.Collect coefficients of $d x$ and $d y$ . This gives $(- x^{3} + 3 x y^{2}) d x + (- 3 x^{2} y + y^{3}) d y = 0$ .
5.Rearrange to obtain exactly $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ . The one arbitrary parameter is consistent with a first-order family on regular solution portions.

Use structure to choose an efficient method

Before applying a formula, inspect products, ratios, and coefficients. Sometimes recognising a product derivative saves work; sometimes an ordinary substitution is helpful. The following examples connect methods instead of grouping problems by a label given in advance.

Example — A product derivative behind a separable relation

Problem
Find the curve through $(0, π /4)$ satisfying $sin x cos y d x + cos x sin y d y = 0$ .

1.The derivative of $cos x cos y$ is the negative of the given expression. Therefore $cos x cos y = C$ .
2.At $(0, π /4)$ , $C = 1/ 2$ . Thus the required local curve is $cos x cos y = 1/ 2$ .
3.Separation would also work where $cos x cos y \neq = 0$ : $tan x d x + tan y d y = 0$ . Recognising the product avoids unnecessary logarithmic rearrangement.

Example — A reciprocal exponential emerges from separation

Problem
Solve $(1 + e^{2 x}) d y + (1 + y^{2}) e^{x} d x = 0$ with $y (0) = 1$ .

1.Separate: $d y / (1 + y^{2}) = - e^{x} d x / (1 + e^{2 x})$ .
2.The substitution $u = e^{x}$ identifies the right integral. Thus $tan^{- 1} y + tan^{- 1} (e^{x}) = C$ .
3.The initial point gives $C = π /2$ . Since $e^{x} > 0$ , $π /2 - tan^{- 1} (e^{x}) = tan^{- 1} (e^{- x})$ .
4.Hence $y = e^{- x}$ . Substituting gives $(1 + e^{2 x}) (- e^{- x}) + (1 + e^{- 2 x}) e^{x} = 0$ , verifying the result.

Example — A ratio substitution beyond the standard appearance

Problem
Solve $y e^{x / y} d x = (x e^{x / y} + y^{2}) d y$ , where $y \neq = 0$ .

1.Solve for $d x / d y$ : $d x / d y = x / y + y e^{- x / y}$ . The extra y factor means this is not itself a degree-zero right-hand side.
2.Nevertheless, the ratio suggests trying $x = v y$ . Then $v + y d v / d y = v + y e^{- v}$ .
3.Since $y \neq = 0$ , cancellation gives $d v / d y = e^{- v}$ . Separate and integrate: $e^{v} d v = d y$ , so $e^{v} = y + C$ .
4.Replace $v$ by $x / y$ : $e^{x / y} = y + C$ . This is a good example of using the reasoning behind a substitution rather than misclassifying an equation.

Example — A difference suggests a new variable

Problem
Solve $(x - y) (d x + d y) = d x - d y$ with $y = - 1$ when $x = 0$ .

1.The repeated difference suggests $t = x - y$ , so $d t = d x - d y$ . Also $d x + d y = 2 d x - d t$ .
2.Substitute to get $t (2 d x - d t) = d t$ . For $t \neq = 0$ , this becomes $2 d x = (1 + 1/ t) d t$ .
3.Integrate: $2 x = t + ln ∣ t ∣ + C$ . Replacing $t = x - y$ gives $x + y - ln ∣ x - y ∣ = C$ .
4.At $(0, - 1)$ , $C = - 1$ . The required local curve is $x + y - ln ∣ x - y ∣ = - 1$ , with $x - y > 0$ near the initial point.
5.The excluded case $t = 0$ gives the line $y = x$ , which satisfies the original relation but not the supplied condition.

Example — Reverse a derivative and then use linear form

Problem
Solve $(\frac{e ^{- 2 x}}{x} - \frac{y}{x}) \frac{d x}{d y} = 1$ , with $x \neq = 0$ .

1.On a portion where the inverse derivative exists, rewrite as $d y / d x = (e^{- 2 x} - y) / x$ .
2.This is $y^{'} + y / x = e^{- 2 x} / x$ , a linear equation with convenient multiplier $μ = x$ .
3.The product equation is $d (x y) / d x = e^{- 2 x}$ . Integration gives $x y = - \frac{1}{2} e^{- 2 x} + C$ .
4.The implicit family must be read on intervals where $x \neq = 0$ and the original inverse derivative exists. At a point where $e^{- 2 x} - y = 0$ , the original $d x / d y$ expression cannot satisfy the displayed equation with a finite derivative.

Example — A particular linear solution in product form

Problem
Solve $y^{'} + y cot x = 4 x csc x$ with $y (π /2) = 0$ .

1.Use the integrating factor $sin x$ on the interval $0 < x < π$ .
2.Multiplying gives $d (y sin x) / d x = 4 x$ . Integrate: $y sin x = 2 x^{2} + C$ .
3.The condition gives $C = - π^{2} /2$ . Therefore $y = (2 x^{2} - π^{2} /2) / sin x$ on that interval.
4.The product form makes both the integration and the condition easy to check.

Example — Exponentiation creates a linear equation

Problem
Solve $(x + 1) y^{'} = 2 e^{- y} - 1$ with $y (0) = 0$ .

1.Multiply by $e^{y}$ : $(x + 1) e^{y} y^{'} = 2 - e^{y}$ . Since $d (e^{y}) / d x = e^{y} y^{'}$ , set $u = e^{y}$ .
2.The equation becomes $(x + 1) u^{'} + u = 2$ . The left side is the derivative of $(x + 1) u$ .
3.Integrate: $(x + 1) u = 2 x + C$ . At $x = 0$ , $u = e^{0} = 1$ , so $C = 1$ .
4.Thus $e^{y} = (2 x + 1) / (x + 1)$ and $y = ln (\frac{2 x + 1}{x + 1})$ .
5.On the solution interval containing zero, positivity of the logarithm argument requires $x > - 1/2$ . This interval also avoids $x = - 1$ .

Example — Reconcile overlapping methods

Problem
Solve $y^{'} = x (1 + y)$ and explain why a missing constant solution would be an error.

1.Separation for $y \neq = - 1$ gives $d y / (1 + y) = x d x$ , hence $ln ∣1 + y ∣ = x^{2} /2 + C_{1}$ .
2.Exponentiation gives $y = - 1 + C e^{x^{2} /2}$ for nonzero C. Substituting into the original equation also permits $C = 0$ , restoring $y = - 1$ .
3.Alternatively, write $y^{'} - x y = x$ , with factor $e^{- x^{2} /2}$ . Integration gives the same family, including the zero value of C.
4.The methods agree. Any apparent difference here comes from division by $1 + y$ , not from different underlying solutions.

The miscellaneous exercise also includes order-and-degree cases, verification of exponential, trigonometric, and implicit logarithmic functions, and inverse-tangent integrations. They belong in mixed review because the reader must decide which kind of task is being asked. The practice below revisits these demands without presorting every question by method.

Make the final check part of the solution

A final formula is only useful if it solves the problem as stated. Differentiate it, substitute it into the original equation, apply the given condition, and inspect the domain. These checks catch sign errors, lost constants, and transformations that have extended a formula beyond its valid interval.

The two source conditions requiring special care illustrate different issues. A cosecant equation cannot accept an initial value making its argument zero. A statement involving the magnitude of a slope cannot require that magnitude to be negative. In both cases, a formal transformed answer is insufficient: the original statement controls validity.

For growth, check units and the quantity described. If $P^{'} = k P$ , the model is $P = P_{0} e^{k t}$ and the doubling time is $ln 2/ k$ . If spherical volume increases at a constant rate, $r^{3}$ is linear in time. These two models describe different rate statements and should not be interchanged.

How the methods are connected historically

The search for curves with prescribed tangents helped shape the early study of differential equations. The historical note traces separation, homogeneous equations, and first-order linear methods to work by Leibniz and the Bernoullis. It also explains why “integral curve” remains a natural name for a solution curve.

The terminology developed over time: James Bernoulli used the word integral in this setting; Lagrange used solution, and Poincare later strongly supported that terminology. Poincare’s portrait at the chapter opening provides context for the later emphasis on understanding how solution curves behave, not only on writing formulas.

The historical example $x^{2} y^{''} = 2 y$ illustrates variety without requiring a new solution method here. On $x \neq = 0$ , $y = x^{2}$ and $y = 1/ x$ each satisfy the equation by direct differentiation, as does their sum $y = x^{2} + 1/ x$ . The first gives a parabola, the second a hyperbola, and the last can be written $x y = x^{3} + 1$ , an algebraic curve of degree three. One differential equation can therefore accommodate quite different-looking curves.

This motivates qualitative analysis: studying the behaviour and shape of solution curves, even when explicit formulas are difficult. For this chapter, the practical lesson is to connect algebra, derivatives, and the meaning of a curve whenever you verify or solve an equation.

Mixed understanding check

These questions cover the whole chapter. Select an answer only after deciding whether you need classification, verification, a method, a constant, or a domain check.

Quick check

For $y^{(4)} - sin (y^{'''}) = 0$ , which is correct?

Quick check

Which check proves that an implicit candidate is a solution?

Quick check

Which method is naturally suggested by $y^{'} = (x + y) / x$ ?

Quick check

Which is a valid integrating factor for $y^{'} + y / x = e^{- 2 x} / x$ ?

Quick check

What is the general solution of $e^{x} d y + (y e^{x} + 2 x) d x = 0$ ?

Quick check

For $(y d x - x d y) / y = 0$ , which standard family is obtained on a suitable nonzero interval?

Quick check

If $P^{'} = k P$ with k positive, what time is required to double a positive initial amount?

Quick check

Which is a particular member of $y = - 1 + C e^{x^{2} /2}$ with $y (0) = 1$ ?

Quick check

What assumption must be checked when writing $ln (y^{'})$ ?

Quick check

For $d x / d y + P_{1} (y) x = Q_{1} (y)$ , which expression is correct?

Practice Problems

Classify order and degree of $y^{''} + 5 x (y^{'})^{2} - 6 y = ln x$ , and state the logarithm restriction.
Verify $x y = a e^{x} + b e^{- x} + x^{2}$ in $x y^{''} + 2 y^{'} - x y + x^{2} - 2 = 0$ for $x \neq = 0$ .
Verify $y = e^{x} (a cos x + b sin x)$ in $y^{''} - 2 y^{'} + 2 y = 0$ .
Verify $y = x sin 3 x$ in $y^{''} + 9 y - 6 cos 3 x = 0$ .
Verify $x^{2} = 2 y^{2} ln y$ in $(x^{2} + y^{2}) y^{'} - x y = 0$ for $y > 0$ .
Solve $y^{'} + (1 - y^{2}) / (1 - x^{2}) = 0$ on valid intervals, and check constant solutions separately.
Integrate $y^{'} + (y^{2} + y + 1) / (x^{2} + x + 1) = 0$ and connect the result to $x + y + 1 = A (1 - x - y - 2 x y)$ .
Find the curve through $(0, π /4)$ satisfying $sin x cos y d x + cos x sin y d y = 0$ . Verify by differentiation.
Solve $y^{'} = (x + 2 y) / (x - y)$ and explain why $y = v x$ is useful.
Solve $y^{'} + y / x = x^{2}$ and select the member through $(1, 2)$ .
Solve $e^{x} d y + (y e^{x} + 2 x) d x = 0$ and identify the product derivative.
Solve $(x + 1) y^{'} = 2 e^{- y} - 1$ with $y (0) = 0$ , stating the interval containing zero.
A quantity grows proportionally to its current value. It increases by 10 percent in two hours. Find its doubling time.
A spherical balloon has constant volume-increase rate, radius 3 at t = 0, and radius 6 at t = 3. Find r at t = 1 and explain why a linear radius model is wrong.
Solve $(x - y) (d x + d y) = d x - d y$ through $(0, - 1)$ , and identify the solution excluded by dividing by x − y.
Compare $y^{'} + sin y = 0$ with $y^{'} + sin (y^{'}) = 0$ : give order, degree, and the reason for the difference.

Order $2$ , degree $1$ , and $x > 0$ .

Key Takeaways

• Classification, verification, and solving are different tasks; identify which one is required. • Separation integrates isolated variable factors; a homogeneous ratio substitution creates separation; an integrating factor creates a product derivative. • Given points and initial values choose particular solutions from families. • Tangent and growth models begin with a careful translation into derivatives and units. • Check original domains, lost constant solutions, and signs after every transformation. • A complete answer explains the method, shows the integration, and verifies the result.

Previous · Lesson 5

Linear Equations and Integrating Factors

End of chapter

See the whole chain of reasoning

Classify and verify before solving

Use structure to choose an efficient method

Make the final check part of the solution

How the methods are connected historically

Mixed understanding check

Quiz

Practice Problems

Key Takeaways