Differential Equations · Lesson 2 of 6

Solutions, Families, and Initial Conditions

“Understand what a solution means, verify it, and see how conditions select a curve from a family.”

Learning Objectives

• Explain why a differential equation has functions as solutions. • Verify explicit and implicit solutions by differentiating and substituting. • Distinguish a general solution from a particular solution. • Use given values to determine arbitrary constants. • Recognise domain and sign restrictions that affect a verification.

A solution is a function

Solving $x^{2} - 1 = 0$ produces the numbers $1$ and $- 1$ . A differential equation asks for something different: a function whose values and derivatives satisfy the equation together. The graph of such a function is called a solution curve, or integral curve.

Definition

Solution

A function is a solution of a differential equation on an interval if the required derivatives exist there and substituting the function and those derivatives makes the equation true throughout that interval.

For $y^{'} = 2 x$ , every function $y = x^{2} + C$ has the required derivative. The constant $C$ determines vertical position, while the differential equation determines the slope at each value of $x$ . There is no need to guess a single privileged curve until an extra condition is supplied.

A family and a selected member— The curves have the same slope at the same horizontal position. Passing through the marked point selects one member.

In the diagram the three curves are $y = x^{2} - 1$ , $y = x^{2}$ , and $y = x^{2} + 1$ . Their derivatives all equal $2 x$ . The condition $y = 2$ when $x = 1$ chooses the third curve because $2 = 1 + C$ gives $C = 1$ .

General and particular solutions

An arbitrary constant can take different values, producing different solutions. A condition on the curve can fix that freedom. These two stages have distinct names, and keeping them separate makes later integration problems easier to organise.

Definition

General solution

A general solution, also called a primitive, describes a family of solutions using arbitrary constants. In the standard general families considered here, the number of independent arbitrary constants equals the order of the differential equation.

Definition

Particular solution

A particular solution is obtained by fixing the arbitrary constants in a general solution. It contains no arbitrary constants, although ordinary numerical constants may remain.

The word “arbitrary” matters. The number $2$ in $y = 2 sin x$ is a constant but is not adjustable; it does not make this a general solution. Also, the labels used for constants are unimportant: $C$ , $K$ , and $C_{1}$ can describe the same freedom if used consistently.

Example — Fix two parameters

Problem
For $y^{''} + y = 0$ , consider the family $y = a sin (x + b)$ . Identify the particular solution obtained when $a = 2$ and $b = \frac{π}{4}$ .

1.The derivative is $y^{'} = a cos (x + b)$ and the second derivative is $y^{''} = - a sin (x + b) = - y$ . Thus every member satisfies the differential equation.
2.Before values are chosen, $a$ and $b$ are arbitrary parameters in the family.
3.Substituting the given values produces $y = 2 sin (x + \frac{π}{4})$ . There are no arbitrary constants left, so this is a particular solution.

Feature	General solution	Particular solution
What it describes	A family of functions	One specified member
Arbitrary constants	Present, with independent freedom	Fixed; none remain arbitrary
First-order illustration	$y = x^{2} + C$	$y = x^{2} + 1$
How to check it	Differentiate and substitute	Differentiate, substitute, and check any given conditions

Verify an explicit solution

Verification is different from finding a solution from scratch. The candidate function is already supplied. Your job is to calculate the derivatives required by the equation and show that the left-hand side equals the right-hand side as an identity, not just at one chosen point.

Example — An exponential solution

Problem
Verify that $y = e^{- 3 x}$ satisfies $y^{''} + y^{'} - 6 y = 0$ .

1.Differentiate once: $y^{'} = - 3 e^{- 3 x}$ . Differentiate again: $y^{''} = 9 e^{- 3 x}$ .
2.Substitute all three expressions into the left-hand side: $9 e^{- 3 x} - 3 e^{- 3 x} - 6 e^{- 3 x}$ .
3.The coefficients sum to $9 - 3 - 6 = 0$ . The left-hand side is therefore zero for every real $x$ , equal to the right-hand side.
4.We have verified this solution without claiming that it is the entire general solution of the second-order equation.

Example — A trigonometric family

Problem
Verify that $y = a cos x + b sin x$ satisfies $y^{''} + y = 0$ .

1.Treat $a$ and $b$ as constants while differentiating: $y^{'} = - a sin x + b cos x$ .
2.Differentiate again: $y^{''} = - a cos x - b sin x$ .
3.Adding $y$ gives $(- a cos x - b sin x) + (a cos x + b sin x) = 0$ .
4.The identity holds for every choice of $a$ and $b$ . This is a two-constant solution family of the second-order equation.

Example — Use a value and a slope

Problem
In the family $y = a cos x + b sin x$ , find the member for which $y (0) = 2$ and $y^{'} (0) = - 1$ .

1.At $x = 0$ , $cos 0 = 1$ and $sin 0 = 0$ , so $y (0) = a = 2$ .
2.The derivative is $y^{'} = - a sin x + b cos x$ , so $y^{'} (0) = b = - 1$ .
3.The required function is $y = 2 cos x - sin x$ . It satisfies both the differential equation already verified and the two given conditions.
4.A single value such as $y (0) = 2$ would not determine $b$ . Two independent pieces of information fix the two constants here.

Example — A product needs the product rule

Problem
Verify that $y = x sin x$ satisfies $x y^{'} = y + x^{2} cos x$ .

1.The product rule gives $y^{'} = sin x + x cos x$ .
2.Multiplying by $x$ gives $x y^{'} = x sin x + x^{2} cos x$ .
3.Since $y = x sin x$ , the right-hand side is exactly $y + x^{2} cos x$ . The identity is verified.

A square root can introduce a sign condition

For $y = x sin x$ , $x^{2} - y^{2} = x^{2} cos^{2} x = ∣ x cos x ∣$ . Therefore the version $x y^{'} = y + x x^{2} - y^{2}$ agrees with the previous identity only on intervals where $x cos x \geq 0$ , with any other stated restrictions also respected. Do not silently replace $u^{2}$ by $u$ for arbitrary real $u$ .

Verify an implicit solution

Sometimes the candidate is given as a relation between $x$ and $y$ , rather than an explicit formula for $y$ . You can still differentiate it by treating $y$ as a function of $x$ . Every differentiation of an expression involving $y$ must then include the chain rule.

Example — An implicit logarithmic relation

Problem
Verify that $x y = ln y + C$ satisfies $y^{'} = \frac{y ^{2}}{1 - x y}$ where $y > 0$ and $x y \neq = 1$ .

1.Differentiate $x y$ by the product rule: $y + x y^{'}$ . Differentiate $ln y + C$ : $\frac{y ^{'}}{y}$ .
2.Thus $y + x y^{'} = \frac{y ^{'}}{y}$ . Multiply by $y$ to obtain $y^{2} + x y y^{'} = y^{'}$ .
3.Collect the derivative terms: $(1 - x y) y^{'} = y^{2}$ . Division by $1 - x y$ yields the required differential equation.
4.The restriction $y > 0$ comes from the logarithm. The restriction $x y \neq = 1$ is needed for the displayed quotient.

Example — A relation involving the inverse tangent

Problem
Verify that $x + y = tan^{- 1} y$ satisfies $y^{2} y^{'} + y^{2} + 1 = 0$ on any differentiable solution interval.

1.Differentiate both sides: $1 + y^{'} = \frac{y ^{'}}{1 + y ^{2}}$ .
2.Multiply by $1 + y^{2}$ : $1 + y^{2} + (1 + y^{2}) y^{'} = y^{'}$ .
3.Subtract $y^{'}$ from both sides to obtain $y^{2} y^{'} + y^{2} + 1 = 0$ .
4.At $y = 0$ , this equation would read $1 = 0$ for a finite derivative. The implicit relation does not define a differentiable solution $y (x)$ through that point; verification is understood on appropriate intervals.

Example — The upper half of a circle

Problem
Verify that $y = a^{2} - x^{2}$ satisfies $x + y y^{'} = 0$ for $a > 0$ and $- a < x < a$ .

1.On this interval $y > 0$ and the derivative exists. By the chain rule, $y^{'} = \frac{- x}{a ^{2} - x ^{2}} = - \frac{x}{y}$ .
2.Substitute into the equation: $x + y (- \frac{x}{y}) = x - x = 0$ .
3.The endpoints $x = \pm a$ are excluded because the derivative formula involves division by zero there. The graph is a solution curve on the stated open interval.

A longer verification organised efficiently

Longer expressions do not require a new meaning of solution. The same differentiate-and-substitute test applies. A useful shorthand can keep repeated product-rule calculations clear without hiding the reasoning.

Example — An exponential–trigonometric family

Problem
Verify $y = e^{a x} (c_{1} cos b x + c_{2} sin b x)$ in $y^{''} - 2 a y^{'} + (a^{2} + b^{2}) y = 0$ .

1.Let $u = c_{1} cos b x + c_{2} sin b x$ . Then $u^{'} = - b c_{1} sin b x + b c_{2} cos b x$ and $u^{''} = - b^{2} u$ .
2.Since $y = e^{a x} u$ , the product rule gives $y^{'} = e^{a x} (a u + u^{'})$ .
3.One more differentiation gives $y^{''} = e^{a x} (a^{2} u + 2 a u^{'} + u^{''})$ .
4.Substitute and factor out $e^{a x}$ . The remaining expression is $a^{2} u + 2 a u^{'} + u^{''} - 2 a (a u + u^{'}) + (a^{2} + b^{2}) u$ .
5.The $u^{'}$ terms cancel, and the $a^{2} u$ terms cancel. What remains is $u^{''} + b^{2} u = 0$ . Hence the proposed function satisfies the equation.
6.This verification works for real $a$ and $b$ . If $b = 0$ , the sine term vanishes; verifying a formula is not the same as proving that its parameters give a full independent general family in every special case.

Checking one point is not verification

A proposed function can satisfy an equation at one point and fail everywhere else. Verify the differential equation as an identity over a valid interval. Then check a supplied value or slope separately to establish the particular solution requested.

Check your understanding

Focus on what is being checked: a derivative identity, an arbitrary constant, or a condition selecting a curve. In implicit equations, remember that differentiating $y$ produces $y^{'}$ .

Quick check

What is a solution of a differential equation?

Quick check

Which is a particular solution of $y^{'} = 2 x$ ?

Quick check

How many independent arbitrary constants are expected in a standard general solution of a fourth-order equation?

Quick check

If $y = x^{2} + C$ and $y (2) = 7$ , what is $C$ ?

Quick check

What is the derivative of $ln y$ with respect to $x$ ?

Quick check

If $y = e^{- 3 x}$ , what is $y^{''}$ ?

Quick check

Which check is sufficient to verify a proposed solution on an interval?

Quick check

How many arbitrary constants remain in a particular solution of a third-order equation?

Practice Problems

Verify $y = e^{x} + 1$ in $y^{''} - y^{'} = 0$ .
Verify $y = x^{2} + 2 x + C$ in $y^{'} - 2 x - 2 = 0$ .
Verify $y = cos x + C$ in $y^{'} + sin x = 0$ .
Verify $y = 1 + x^{2}$ in $y^{'} = \frac{x y}{1 + x ^{2}}$ .
Verify $y = A x$ in $x y^{'} = y$ for $x \neq = 0$ , and identify whether A remains arbitrary.
For $y = x sin x$ , explain why replacing $x^{2} cos^{2} x$ by $x cos x$ needs a sign check.
Verify $y - cos y = x$ in $(y sin y + cos y + x) y^{'} = y$ .
Find the member of $y = x^{2} + C$ passing through $(- 1, 4)$ .
Find the member of $y = a cos x + b sin x$ with $y (0) = 0$ and $y^{'} (0) = 3$ .
Explain why $y = e^{- 3 x}$ is a verified solution but is not by itself a two-constant general solution.
Verify $x y = ln y + C$ in $(1 - x y) y^{'} = y^{2}$ , stating the logarithm restriction.
For $x + y = tan^{- 1} y$ , differentiate and obtain the polynomial derivative relation.

$y^{'} = e^{x}$ and $y^{''} = e^{x}$ , so the difference is zero.

Key Takeaways

• A solution is a function that satisfies the differential equation over an appropriate interval. • Verify by differentiating and substituting; use implicit differentiation when necessary. • General solutions contain independent arbitrary constants; particular solutions have those constants fixed. • A given value or slope supplies information for choosing a member of a family. • Domains and sign conditions are part of a correct verification.

Previous · Lesson 1

Recognising Differential Equations

Next · Lesson 3

Separation of Variables and Growth Models

A solution is a function

General and particular solutions

Verify an explicit solution

Verify an implicit solution

A longer verification organised efficiently

Check your understanding

Quiz

Practice Problems

Key Takeaways