A unified route from a family of curves to separable, ratio-homogeneous and linear first-order equations.
JEE Main · JEE Advanced100 min study guide14 indexed sections
BY THE END, YOU CAN
Form differential equations by eliminating genuinely independent constants.
Solve separable equations while restoring equilibrium solutions lost by division.
Recognize first-order ratio-homogeneous equations and substitute y = vx correctly.
Derive and use integrating factors for first-order linear equations.
Apply initial conditions and state an interval where the original equation and solution both make sense.
BEFORE YOU START
Differentiation, product and chain rules.
Indefinite integration, logarithms, exponential functions and partial fractions.
Implicit curves and interval domains.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2021. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Matched topics: Linear Differential Equations.
JEE Advanced · 2021 · Paper 2 · Mathematics · Q4
Paraphrased exam question
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
For real constants α,β, require y′+αy=xe^(βx) and y(1)=1 on ℝ. Which functions qualify? A=(x²/2+e−1/2)e⁻ˣ; B=(−x²/2+e+1/2)e⁻ˣ; C=(x/2−1/4)eˣ+(e−e²/4)e⁻ˣ; D=(1/4−x/2)eˣ+(e+e²/4)e⁻ˣ.
Reveal answer & worked solution
Answer: A and C (independently derived).
Every listed function has y(1)=1. At x=0 each has y′(0)=−y(0) and y(0)≠0, so the differential equation forces α=1.
Differentiate and add y: A gives xe⁻ˣ, B gives −xe⁻ˣ, C gives xeˣ, and D gives −xeˣ.
Thus A works with β=−1 and C with β=1. B and D cannot equal xe^(βx) for real β: for x>0 their left sides are negative whereas the required right side is positive.
Source: JEE (Advanced) 2021, Paper 2, Mathematics Q4; independently derived solution; key finality unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. An equation for a function rather than a number
An algebraic equation asks for unknown numbers. A differential equation asks for an unknown function whose values and derivatives obey a rule on an interval. In y′ = 2x, every function y = x² + C works. The constant labels different curves. A condition such as y(1) = 4 selects the curve with C = 3. A solution must satisfy the rule throughout its stated interval, not merely at the initial point.
The order is the highest derivative present. A first-order equation contains y′ but no higher derivatives; a second-order equation may contain y″. Linearity is a separate property: y′ + P(x)y = Q(x) is linear in the unknown function and its derivative even if P or Q is complicated. In contrast, y′ = y² and yy′ = x are nonlinear. The independent variable may appear nonlinearly without spoiling linearity in y.
This lesson treats four connected skills: forming an equation from a curve family, solving separable equations, converting ratio-homogeneous equations by a substitution, and solving linear equations by an integrating factor. A single equation may belong to more than one category. Choose a method that preserves the domain and makes the algebra transparent rather than forcing every problem into one memorized form.
For a first-order initial-value problem y′ = F(x,y), y(x₀) = y₀, the usual smoothness conditions near (x₀,y₀) guarantee a local unique solution. We do not assume these conditions automatically. A formula that blows up later is still a local solution; a point where the original equation is undefined cannot be crossed merely because a rearranged expression appears finite.
02 / UNDERSTAND & APPLY
2. Forming an equation: eliminate the labels of a family
A curve family y = f(x;C₁,…,Cₙ) contains constants that do not change as x changes along one member. Differentiate enough times to obtain relationships involving those constants, then eliminate them. The resulting differential equation is a common rule satisfied by all family members. Do not eliminate the independent variable x or the unknown function y; the quantities to eliminate are the arbitrary constants.
The number of independent constants suggests the expected order, but counting written letters mechanically is unsafe. In y = (A + B)eˣ, only the combination A + B matters, so the family has one effective constant and yields a first-order equation. Similarly, differentiating more times than necessary can produce a higher-order equation whose solution family is much larger than the original one.
An equation obtained by differentiation may admit extra solutions or lose information about a restricted family. For example, eliminating a positive parameter may remove its positivity condition. Record such restrictions if the problem asks for exact equivalence rather than merely an equation satisfied by the family. If elimination involves squaring or division, check what assumptions were introduced.
The degree of a differential equation, when defined, is the power of its highest-order derivative after the equation is written polynomially in derivatives with the relevant algebraic simplifications. It is not the order. An expression involving sin(y′) is not polynomial in derivatives, so assigning degree one just because y′ is first-order is incorrect. Formation problems should primarily be checked by substitution into the resulting differential equation.
ORIGINAL WORKED EXAMPLE
Two truly independent constants
Form a differential equation for y = A eˣ + B e⁻ˣ, where A and B are arbitrary real constants.
Differentiate once: y′ = A eˣ − B e⁻ˣ. Differentiate again: y″ = A eˣ + B e⁻ˣ.
The second derivative equals the original expression y, so eliminating both constants gives y″ − y = 0.
This is second-order and linear. Substitution of any member A eˣ + B e⁻ˣ makes the left side zero.
The constants are genuinely independent because eˣ and e⁻ˣ are not constant multiples of one another. The usual two-constant solution family matches the second-order equation.
Answer y″ − y = 0.
ORIGINAL WORKED EXAMPLE
A one-parameter family of circles
Form a differential equation satisfied by x² + y² = 2Cx on intervals where a circle branch can be written as differentiable y(x).
Treat C as constant and differentiate: 2x + 2yy′ = 2C, hence C = x + yy′.
Substitute into the original relation: x² + y² = 2x(x + yy′).
Rearrange to obtain 2xyy′ = y² − x². This expression has no arbitrary parameter.
The derivation applies to differentiable graph branches. A circle may have a vertical tangent where y(x) is not differentiable, so the graph form does not describe that tangent as a finite y′.
Answer 2xyy′ = y² − x², on differentiable graph branches.
03 / UNDERSTAND & APPLY
3. Separable equations: put each variable on its own side
An equation is separable if it can be written y′ = g(x)h(y). On a region where h(y) ≠ 0, rearrange to dy/h(y) = g(x)dx and integrate. The notation with differentials is a convenient form of the chain rule: if H′(y) = 1/h(y), then d[H(y(x))]/dx = y′/h(y) = g(x). Thus H(y) = G(x) + C.
Before dividing by h(y), solve h(y) = 0. Every real constant y = c satisfying h(c) = 0 gives a potential equilibrium solution, provided the original equation is defined there. Such a constant solution disappears from the divided equation. List and check these equilibria separately, even if some are later recovered by extending a constant in an explicit formula.
Only one arbitrary integration constant is needed after integrating both sides: the difference of two arbitrary constants is another arbitrary constant. When integrating 1/y, use ln|y| on intervals where y ≠ 0. Exponentiation yields a nonzero signed multiplicative constant, not only a positive one, until any separately checked zero solution is restored.
An implicit relation may be the clearest final answer. It defines a differentiable solution locally only where it can actually be solved for y with the required regularity. A squared expression may describe more than one branch; an initial value and continuity choose the appropriate one. Never erase that branch choice by writing an unexplained ± sign.
y′ = g(x)h(y) ⇒ ∫dy/h(y) = ∫g(x)dx + C
First inspect h(y) = 0; the divided equation only handles non-equilibrium regions.
ORIGINAL WORKED EXAMPLE
Separation with an initial condition
Solve y′ = 2xy with y(0) = 3.
The original equation admits the equilibrium y = 0, but it does not satisfy the given initial condition.
For the requested nonzero solution, divide by y and integrate: ln|y| = x² + C.
Exponentiation gives y = A e^(x²) with nonzero signed A. Since y(0) = 3, A = 3.
Check: y′ = 6x e^(x²) = 2x(3e^(x²)). The solution is defined for every real x and satisfies the initial value.
Answer y = 3e^(x²), for all real x.
04 / UNDERSTAND & APPLY
4. Equilibria and an instructive partial-fraction example
An autonomous equation y′ = h(y) has a slope depending only on the current value y. Its equilibrium values divide the y-axis into regions where the derivative has a fixed sign. This qualitative picture can predict whether a solution rises or falls before you perform an integral. It also checks whether an explicit answer behaves sensibly.
For y′ = y(1 − y), the equilibria are 0 and 1. Between them the derivative is positive, so a solution rises. Above 1 the derivative is negative, so it falls. Below 0 the derivative is negative as well. The formula may have a finite-time pole in some regions, so a single algebraic expression must be restricted to an interval that avoids its denominator zeros.
Partial fractions give 1/[y(1 − y)] = 1/y + 1/(1 − y). The second term integrates to −ln|1 − y|, not +ln|1 − y|. The logarithmic relation ln|y/(1 − y)| = x + C follows on non-equilibrium branches. The signs in this step matter: a wrong sign produces a different equation even if the result looks like a familiar growth formula.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
This is the particular solution branch of the nonlinear separable example inside the broader differential-equations lesson. It is not presented as a linear differential equation or as a time model. The displayed window is finite; the solution domain is all real x.
ORIGINAL WORKED EXAMPLE
A logistic initial-value problem
Solve y′ = y(1 − y), y(0) = 1/4, and describe its behavior.
Check the equilibria y = 0 and y = 1. Neither has the required initial value, so separate on a region with y ≠ 0,1.
Integrate using partial fractions to obtain ln|y/(1 − y)| = x + C. The initial value lies between 0 and 1, so the ratio is positive on its solution branch.
Write y/(1 − y) = K eˣ. At x = 0 the ratio is (1/4)/(3/4) = 1/3, so K = 1/3.
Solve for y: y = eˣ/(3 + eˣ) = 1/(1 + 3e⁻ˣ). The denominator is positive for all real x.
The solution stays between 0 and 1, increases, tends to 0 as x tends to −∞, and tends to 1 as x tends to +∞. Direct differentiation gives y′ = 3e⁻ˣ/(1 + 3e⁻ˣ)² = y(1 − y).
Answer y = 1/(1 + 3e⁻ˣ), on ℝ; increasing from the limiting value 0 toward the limiting value 1.
05 / UNDERSTAND & APPLY
5. Initial values select branches and intervals
Consider y′ = x/y. Integration produces y² = x² + C, but the original equation forbids y = 0. An initial value with y > 0 chooses the positive square-root branch and one with y < 0 chooses the negative branch. If the radicand vanishes at a finite x, the solution interval ends there for the original explicit equation, even though the implicit algebraic curve itself continues.
The maximal interval for an initial-value solution is a connected interval containing the initial x-value on which the function is differentiable and the original equation is satisfied. It cannot jump across a pole. For y′ = y² and y(0) = 1, the formula y = 1/(1 − x) is algebraically defined on two components, but the initial-value solution has maximal interval (−∞,1), the component containing zero.
Uniqueness also deserves care. For y′ = 2√y with y(0) = 0, the right side is defined for y ≥ 0 but is not locally Lipschitz in y at zero. There are solutions that remain zero until a chosen time and then follow a squared branch. A division by √y would miss such behavior. In routine smooth examples, use the standard existence-and-uniqueness conditions; do not promote those conclusions to every separable equation.
ORIGINAL WORKED EXAMPLE
A finite-time endpoint
Solve y′ = y² with y(0) = 2 and identify the maximal interval containing zero.
The equilibrium y = 0 does not meet the initial condition. For y ≠ 0, separate y⁻²dy = dx.
Integration gives −1/y = x + C. At x = 0 and y = 2, C = −1/2.
Hence y = 2/(1 − 2x). Its denominator vanishes at x = 1/2.
The solution is differentiable and satisfies y′ = 4/(1 − 2x)² = y² on the component containing zero, namely (−∞,1/2). It cannot be continued through the blow-up as a finite differentiable solution.
Answer y = 2/(1 − 2x), for −∞ < x < 1/2.
06 / UNDERSTAND & APPLY
6. Ratio-homogeneous equations and the substitution y = vx
In the first-order substitution method, homogeneous means an equation can be written y′ = F(y/x) on a region with x ≠ 0. Equivalently, in a form M(x,y)dx + N(x,y)dy = 0, the functions M and N may be homogeneous of the same degree, with the appropriate denominators nonzero. Scaling both x and y by the same nonzero factor preserves the ratio y/x and hence the direction field.
This meaning is different from homogeneous linear, which means a linear equation has zero forcing term Q(x) = 0. An equation can be homogeneous in one sense and not the other. Always identify which meaning you are using. For instance, y′ = 1 + y/x is ratio-homogeneous but is a nonhomogeneous linear equation when written y′ − y/x = 1.
Set v = y/x, so y = vx. Because v is now an unknown function of x, the product rule gives y′ = v + xv′. Substitution yields xv′ = F(v) − v. This is separable where x ≠ 0 and F(v) − v ≠ 0. The extra v term is essential; writing y′ = xv′ alone destroys the equation.
Before dividing by F(v) − v, solve F(v) = v. Each admissible constant v = k gives the straight-line solution y = kx on an interval where the original equation is defined. These are the analogue of equilibrium solutions in the transformed variable. Restore them explicitly when the divided integral omits them.
y = vx ⇒ y′ = v + x dv/dx
The product rule turns y′ = F(y/x) into x dv/dx = F(v) − v.
ORIGINAL WORKED EXAMPLE
The simplest ratio substitution
Solve y′ = 1 + y/x on an interval not containing zero.
Put y = vx. Then v + xv′ = 1 + v, so xv′ = 1.
Integrate v′ = 1/x to obtain v = ln|x| + C.
Restore y = xv: y = x(ln|x| + C). This formula is valid separately on positive or negative x-intervals.
Differentiate the result: y′ = ln|x| + C + 1, which equals 1 + y/x. There are no constant-v solutions because F(v) − v = 1 never vanishes.
Answer y = x(ln|x| + C), on an interval in x > 0 or x < 0.
07 / UNDERSTAND & APPLY
7. A homogeneous equation with a lost straight-line solution
The homogeneous substitution is not a license to remove every denominator and forget its origin. The ratio y/x requires x ≠ 0, and the original F may exclude other values such as y = 0. A line y = kx found from F(k) = k is valid only when that ratio lies in the original domain.
After integration, constants inside logarithms or reciprocal expressions can hide special solutions. Rather than trying to force every special case into one formula, state the generic family and the exceptional lines separately. This is particularly clean when an omitted line corresponds to division by v² or by a polynomial in v.
A geometric sketch can clarify the role of these lines. Draw the admissible constant-ratio rays from the origin and a few sample direction segments. Since the slope depends only on y/x, all points on one ray share a slope. The solution curves need not themselves be rays; only the values satisfying F(k) = k produce straight-line solutions.
ORIGINAL WORKED EXAMPLE
A missing solution recovered
Solve y′ = y/x + (y/x)² for x ≠ 0, including any solution lost during separation.
Put y = vx and differentiate: v + xv′ = v + v², so xv′ = v².
The constant solution v = 0 gives y = 0, which is valid in the original equation on any interval avoiding x = 0.
For v ≠ 0, separate dv/v² = dx/x. Integrate: −1/v = ln|x| + C, or v = 1/(K − ln|x|).
Thus y = x/(K − ln|x|), with x ≠ 0 and K − ln|x| ≠ 0. Each connected interval must avoid both kinds of excluded point.
Differentiating y = x/D with D = K − ln|x| gives y′ = 1/D + 1/D², exactly y/x + (y/x)².
Answer y = 0, or y = x/(K − ln|x|), on admissible connected intervals.
08 / UNDERSTAND & APPLY
8. Why a linear integrating factor works
For y′ + P(x)y = Q(x), the obstacle is that the left side is almost a product derivative. Multiplying by a nonzero function μ(x) gives μy′ + μPy. To make this equal (μy)′ = μy′ + μ′y, require μ′ = Pμ. Solving this separable equation gives μ = exp(∫P dx). Any nonzero constant multiple is equally useful, so no arbitrary constant is needed in the chosen integrating factor.
On an interval where P and Q are continuous, μ can be chosen positive and differentiable. The equation becomes (μy)′ = μQ, and integration gives μy = ∫μQ dx + C. The arbitrary constant belongs in this final integral. This derivation explains both the formula and why dividing by the coefficient of y′ must happen before reading off P.
If the equation starts as a(x)y′ + b(x)y = c(x), first choose an interval where a(x) ≠ 0 and divide by a(x). Then P = b/a and Q = c/a. A zero of a can be a singular point for the normalized equation. It may require separate analysis in the original equation; it must not disappear from the domain discussion.
The integrating factor is a method for all first-order linear equations with suitable coefficients, not just those with constant P. A solution is the sum of one particular solution and a constant multiple of exp(−∫P dx), the homogeneous linear solution. This superposition statement belongs to linear equations and generally fails for nonlinear ones.
μ(x) = exp(∫P(x)dx), (μy)′ = μQ
After normalization to y′ + Py = Q, choose any nonzero constant multiple of μ.
y = [∫μQ dx + C]/μ
Use a connected interval on which the coefficients and chosen integrating factor are valid.
09 / UNDERSTAND & APPLY
9. Constant and variable coefficients
A constant P often yields an exponential integrating factor. A term proportional to 1/x yields a power of |x|, with the sign interval fixed in advance. If an integrating factor differs by a nonzero constant from the exponential formula, it is still valid. For example, on x < 0, μ = x can serve when P = 1/x even though exp(ln|x|) = |x|, because x = −|x| on that interval.
After obtaining the general solution, apply the initial value to the complete expression. Applying it before integrating may be convenient with definite-integral notation, but simply deleting C because an initial condition exists is incorrect. Finally differentiate and substitute into the original, unnormalized equation when possible; this checks both the algebra and any hidden coefficient restrictions.
ORIGINAL WORKED EXAMPLE
An exponential integrating factor
Solve y′ + 2y = 6 with y(0) = 1.
The equation is already normalized, with P = 2 and Q = 6. Choose μ = e^(2x).
Multiply to obtain (e^(2x)y)′ = 6e^(2x). Integrate: e^(2x)y = 3e^(2x) + C.
Thus y = 3 + Ce^(−2x). The initial value gives 1 = 3 + C, hence C = −2.
Check y′ = 4e^(−2x), and y′ + 2y = 4e^(−2x) + 6 − 4e^(−2x) = 6. All coefficients are continuous on ℝ.
Answer y = 3 − 2e^(−2x), on ℝ.
ORIGINAL WORKED EXAMPLE
Normalization is not optional
Solve xy′ + 2y = x² with y(1) = 2.
The initial point lies in x > 0. Divide by x on that interval: y′ + (2/x)y = x.
Choose μ = exp(∫2/x dx) = x². Then (x²y)′ = x³.
Integration gives x²y = x⁴/4 + C, so y = x²/4 + C/x².
The condition y(1) = 2 gives C = 7/4. Thus y = x²/4 + 7/(4x²). Its pole at zero prevents continuation through zero for this initial-value solution.
Substitution in the original equation gives xy′ + 2y = x² after the reciprocal-square terms cancel.
Answer y = x²/4 + 7/(4x²), for x > 0.
10 / UNDERSTAND & APPLY
10. Choose the dependent variable deliberately
Some equations are not linear in y as a function of x but become linear in x as a function of y. For example, dx/dy + A(y)x = B(y) is a first-order linear equation with independent variable y. Its integrating factor is exp(∫A(y)dy), and every integration must use dy. Switching variables requires a local branch where the inversion or differential formulation is valid.
A useful classification order is: inspect equilibria and obvious restrictions; test whether variables separate; test for a linear form in either dependent variable; test whether the slope depends only on a ratio after an allowed substitution. This is a practical decision process rather than a theorem that every first-order equation fits one of these categories. Some do not have elementary closed-form solutions.
An equation such as y′ = 1 + y/x can be solved either by the homogeneous substitution or as y′ − y/x = 1 with integrating factor 1/x on a sign interval. Agreement between the methods is a useful check. By contrast, y′ + y² = x is neither linear in y nor obviously separable; calling the coefficient of y² “P” does not make the integrating-factor method valid.
ORIGINAL WORKED EXAMPLE
Linear with x as the unknown function
Solve dx/dy + x/y = y for y > 0, with x = 2 when y = 1.
Read the independent variable as y. The coefficient of x is 1/y, so the integrating factor is μ(y) = y.
Multiply: y dx/dy + x = y², which is d(yx)/dy = y².
Integrate to obtain yx = y³/3 + C and hence x = y²/3 + C/y.
At y = 1, x = 2 gives C = 5/3. The solution is x = y²/3 + 5/(3y), on y > 0.
This is a valid x(y) relation even where it may not be globally one-to-one as a function of y. Solving globally for y(x) would require an additional branch analysis.
Answer x = y²/3 + 5/(3y), for y > 0.
11 / UNDERSTAND & APPLY
11. Original practice: formation and separation
Each solution below includes the eliminated constants or lost equilibria that determine whether the method is complete. These are original exercises, with no claimed past-paper origin.
ORIGINAL PRACTICE · 1
Form a differential equation for y = Ax², where A is arbitrary.
Reveal answer & reasoning
xy′ − 2y = 0.
Differentiation gives y′ = 2Ax. Multiplying by x yields xy′ = 2Ax² = 2y, eliminating A without dividing by x. Every family member satisfies the resulting equation, including at zero.
ORIGINAL PRACTICE · 2
Form an equation for y = A cos x + B sin x.
Reveal answer & reasoning
y″ + y = 0.
Differentiate twice: y″ = −A cos x − B sin x = −y. The two constants are removed. Substitution confirms the complete family satisfies this second-order relation.
ORIGINAL PRACTICE · 3
How many effective arbitrary constants occur in y = (A − B)x³, and what first-order equation does it satisfy?
Reveal answer & reasoning
One effective constant; xy′ − 3y = 0.
Only C = A − B affects the function. Then y = Cx³ and y′ = 3Cx², so xy′ = 3y. Counting the written symbols A and B as independent curve parameters would overstate the needed order.
ORIGINAL PRACTICE · 4
Solve y′ = 3x²y, y(0) = −2.
Reveal answer & reasoning
y = −2e^(x³), on ℝ.
For nonzero y, ln|y| = x³ + C. The signed multiplicative constant is −2, not +2. The equilibrium y = 0 exists but does not satisfy this initial value.
ORIGINAL PRACTICE · 5
Solve y′ = x/y, y(0) = 2.
Reveal answer & reasoning
y = √(x² + 4), on ℝ.
Integrating y dy = x dx gives y² = x² + C. The initial condition gives C = 4 and selects the positive branch. The result never reaches the excluded value y = 0.
ORIGINAL PRACTICE · 6
List all constant solutions of y′ = (y − 2)(y + 1).
Reveal answer & reasoning
y = 2 and y = −1.
A constant solution has derivative zero, so the right side must vanish. Both constants are allowed by the polynomial right side. Both would be lost if one divided by the two factors without recording them.
ORIGINAL PRACTICE · 7
Solve y′ = −y², y(0) = 1, with its maximal interval.
Reveal answer & reasoning
y = 1/(x + 1), on (−1,∞).
Separate to obtain −1/y = −x + C. The initial value gives y = 1/(x + 1). The denominator vanishes at −1; the interval containing zero lies to its right.
ORIGINAL PRACTICE · 8
Solve y′ = (1 + y²)/(1 + x²), y(0) = 0.
Reveal answer & reasoning
y = x, on ℝ.
Separation gives arctan y = arctan x + C locally. The initial value gives C = 0. Therefore y = x, and direct substitution verifies the formula globally. There are no real equilibria because 1 + y² never vanishes.
12 / UNDERSTAND & APPLY
12. Original practice: substitutions and integrating factors
When checking a homogeneous equation, explicitly write v + xv′. When checking a linear equation, display its normalized P and integrating factor. These intermediate lines are where most avoidable errors occur.
ORIGINAL PRACTICE · 1
Solve y′ = y/x on an interval avoiding x = 0.
Reveal answer & reasoning
y = Cx.
With y = vx, the equation becomes v + xv′ = v, so v′ = 0 and v is any constant. Dividing by F(v) − v would be impossible because it is identically zero; handle this directly.
ORIGINAL PRACTICE · 2
Solve y′ = 2y/x with y(1) = 3.
Reveal answer & reasoning
y = 3x², for x > 0 as an initial-value interval.
With y = vx, xv′ = v. Nonzero v gives ln|v| = ln|x| + C, hence y = Ax². The lost v = 0 produces the included A = 0 case. The initial value selects A = 3; the original coefficient is undefined at zero.
ORIGINAL PRACTICE · 3
Solve y′ = (x² + y²)/(xy), and identify its excluded values.
Reveal answer & reasoning
y² = x²(2 ln|x| + C), locally on branches with x ≠ 0 and y ≠ 0.
Set v = y/x. Then v + xv′ = 1/v + v, so v dv = dx/x. Integration gives v² = 2 ln|x| + C. The original equation excludes x = 0 and y = 0; real differentiable branches need positive radicand and a fixed sign for y.
ORIGINAL PRACTICE · 4
Is y′ + y = x a homogeneous equation in the ratio sense or in the linear zero-forcing sense?
Reveal answer & reasoning
Neither.
Its slope x − y is not solely a function of y/x, and its linear forcing term is Q(x) = x, which is not identically zero. It remains a first-order linear equation.
ORIGINAL PRACTICE · 5
Solve y′ − y = 0, y(0) = 4.
Reveal answer & reasoning
y = 4eˣ, on ℝ.
Here P = −1 and μ = e⁻ˣ. The equation becomes (e⁻ˣy)′ = 0, so y = Ceˣ. The initial value gives C = 4.
ORIGINAL PRACTICE · 6
Solve y′ + y/x = 3x for x > 0, with y(1) = 2.
Reveal answer & reasoning
y = x² + 1/x.
The integrating factor is x. Then (xy)′ = 3x², so xy = x³ + C. The initial condition gives C = 1, and the solution is valid on the positive interval.
ORIGINAL PRACTICE · 7
Solve y′ + 2xy = 2x with y(0) = 3.
Reveal answer & reasoning
y = 1 + 2e^(−x²), on ℝ.
Choose μ = e^(x²). Then (μy)′ = 2xe^(x²), whose antiderivative is e^(x²). Thus y = 1 + Ce^(−x²); the initial value gives C = 2.
ORIGINAL PRACTICE · 8
Solve dx/dy − 2x/y = y² for y > 0, with x(1) = 0.
Reveal answer & reasoning
x = y³ − y².
The integrating factor in variable y is y⁻². Hence d(x/y²)/dy = 1, giving x/y² = y + C. The condition x(1) = 0 gives C = −1. Direct differentiation confirms dx/dy − 2x/y = y².
13 / UNDERSTAND & APPLY
13. Verify before declaring a solution
A strong verification substitutes the proposed function and its derivative into the original equation, simplifies to an identity, and checks the initial condition separately. For an implicit relation, differentiate it and recover the original equation on the stated region. Do not merely re-run the same integration, because the same algebraic error may survive both passes.
Next inspect every division made during the solution. Dividing by y may remove y = 0; dividing by F(v) − v may remove a line; dividing by x may restrict the independent-variable interval. If a factor could vanish, return to the original equation at that case. Some vanished factors give real special solutions, while others mark forbidden points. The original equation decides.
Finally distinguish a family expression from an initial-value branch. A formula with a pole can define separate solutions on separate intervals, but a single initial-value solution uses the connected interval containing its initial point. An implicit equation with two square-root branches needs a sign choice. A logarithm imposes a nonzero argument before absolute values; absolute values make the real logarithm valid on either sign interval, not at zero.
Formation moves from curve families to derivative relations by eliminating constants. Separation moves from y′ = g(x)h(y) to antiderivatives after recording equilibria. Ratio homogeneity moves from y′ = F(y/x) to a separable equation in v = y/x. Linearity moves from y′ + Py = Q to a product derivative using μ = exp(∫P dx). Each method has its own structural reason; memorized surface appearance is less reliable.
The common wrong moves are differentiating y = vx as though v were constant, forgetting the absolute value in a logarithmic antiderivative, treating e^C as a signed arbitrary constant without restoring the sign, reading P before normalizing the y′ coefficient, and assuming every initial value guarantees one global solution. Correcting these habits matters more than collecting more formula patterns.
For a compact exam solution, write the classification and domain first, show the decisive transformed equation, integrate with one arbitrary constant, restore exceptional solutions, apply any initial condition, and state the final interval. A final derivative substitution is often only one line and provides a disproportionately valuable check.
KEEP THE BIG IDEAS
Quick revision
Eliminate effective constants, not merely symbols, when forming an equation.
Check equilibrium solutions before division in a separable equation.
For y = vx, the derivative is v + xv′; restore constant-ratio lines.
Normalize linear equations before choosing μ = exp(∫P dx).
A solution includes its valid interval and any branch selected by the initial condition.
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