First-order equations and solution methods

Choose a structure and record what a transformation can change

A first-order equation may admit several solution methods, but every method has a domain. Start by writing the unknown function, independent variable, original equation and initial data. Note where coefficients, denominators and real powers are defined. Then identify a useful structure: a product f(t)g(y), a linear expression y′+p(t)y=q(t), an exact differential relation, or a substitution that produces one of these forms.

The methods below derive equivalent relations only where their operations are legitimate. Dividing by g(y) can remove equilibrium solutions. Dividing by a coefficient of y′ excludes its zeros. Taking a logarithm requires a nonzero argument with the appropriate absolute value, and solving for a real root selects branches. Maintain a list of these restrictions and check the excluded cases directly in the original equation.

Lesson 137 established what an initial-value solution means, along with local existence and uniqueness conditions. Here those ideas guide algebraic solution methods. A formula is finally accepted only after differentiation, substitution, the initial-data check and a connected interval containing the initial time. If an equation has no convenient elementary integral, a definite-integral or implicit answer can still be exact.

The examples use dimensionless x or t unless physical units are stated. For physical models, coefficients and exponential arguments must be dimensionally consistent. No finite list of pattern-matching rules solves every nonlinear first-order equation; a failed pattern recognition is a reason to use other analysis, not a reason to alter the equation until it fits.

Separation is a chain-rule argument

Consider y′=a(t)b(y), with a continuous and b continuous on the relevant intervals. On an interval of a solution where b(y(t))≠0, choose an antiderivative G with G′(u)=1/b(u). The chain rule gives dG(y(t))/dt=y′(t)/b(y(t))=a(t). Thus G(y(t))−G(y₀)=∫_(t₀)^t a(s) ds, provided the initial state belongs to that same nonzero branch.

The familiar notation dy/b(y)=a(t) dt abbreviates this argument. It is useful bookkeeping, but it does not turn arbitrary symbols into independent quantities that may always be divided or cancelled. The nonzero condition on b is what makes G′ meaningful. The definite-integral version retains the initial value directly and avoids carrying two unnecessary integration constants.

Conversely, if a differentiable branch satisfies G(y(t))−G(y₀)=∫a and b is defined and nonzero along it, differentiation recovers y′=a(t)b(y). If G is locally invertible there, one can solve for y locally. A global inverse, a convenient elementary antiderivative and an explicit formula are different additional conveniences, not automatic consequences.

Before dividing, solve b(c)=0. Each admissible constant function y=c is an equilibrium solution of the original equation, regardless of a(t). These solutions are not represented by a separated expression containing 1/b(c). A final family that happens to extend to an equilibrium at a limiting constant may include it, but that inclusion must be checked rather than assumed.

Dividing out an equilibrium can hide the requested solution

For y′=y(1−y), the equilibria are 0 and 1. On a branch avoiding both values, partial fractions allow separation by 1/[y(1−y)]. If the initial value is y(0)=0, that divided equation is unavailable at the initial point, whereas y=0 solves the original problem immediately. A method should preserve that simple answer rather than declare the initial-value problem unsolvable.

When the original field is locally Lipschitz in the state, a nonconstant solution cannot meet an equilibrium solution at a finite point in its interval and then depart differently. Local uniqueness at a meeting point would force local agreement, and continuation of that agreement over the common interval prevents a crossing. Polynomial logistic fields have this local regularity. The claim concerns the original equation and its domain.

Without uniqueness, equilibrium waiting and departure can occur. The delayed cubic examples for y′=3|y|^(2/3) in lesson 137 satisfy the equation while spending an interval at zero. Dividing by |y|^(2/3) loses their zero pieces and cannot describe all such histories. An implicit separated branch alone does not settle the entire non-Lipschitz initial-value problem.

This distinction helps organise a solution: list equilibria first, solve nonzero branches second, and use the actual uniqueness hypotheses to describe how branches can meet. Do not infer non-crossing only from a plotted family. Conversely, do not add arbitrary spliced curves to a smooth unique problem merely because each piece separately satisfies a transformed relation away from its join.

Logarithms preserve signs through a branch, not by erasing them

For y′=a(t)y on a nonzero branch, separation gives ln|y(t)|−ln|y₀|=∫_(t₀)^t a(s) ds. Exponentiating yields y(t)=y₀ exp(∫a), since the branch retains its initial sign. The exponential factor is positive, so positive and negative nonzero initial values remain of their respective signs while the integral exists.

In an indefinite calculation, ln|y|=A(t)+C gives |y|=e^C e^(A(t)), with e^C>0. Restoring either sign produces y=C₁e^(A(t)) with C₁≠0. The equilibrium y=0 was excluded in the division and can be appended after direct verification, allowing the final constant C₁ to range over all real numbers in this particular equation.

An absolute value cannot be dropped silently when negative states are allowed. For example ∫dy/y=ln|y|+C on either half-line. A logarithm of a physical dimensional quantity should also be understood through a reference ratio: ln|y/y₀| is dimensionless when y and y₀ share units. The definite expression displays that ratio naturally.

After solving an algebraic relation, inspect denominators and real-valued branches. A squared relation may contain more than one sign choice, and an initial value fixes the local choice only where it can be continued differentiably. Constants are not permission to glue disconnected intervals or cross a forbidden state. Verification in the original equation remains the final algebraic safeguard.

The logistic formula has equilibria and interval cases

For N′=rN(1−N/K), with r,K>0, first record N=0 and N=K. On a branch with neither value, the identity 1/[N(1−N/K)]=1/N+1/(K−N) gives ln|N/(K−N)|=r(t−t₀)+C. Solving and applying N(t₀)=N₀≠0 gives N(t)=K/[1+A e^(−r(t−t₀))], where A=K/N₀−1.

If 0<N₀<K, then A>0, the denominator stays positive for all real time, and N increases toward K in forward time. For N₀=K, A=0 gives the equilibrium formula; N₀=0 must be retained separately. These cases show precisely which equilibrium the divided derivation omitted and how the final formula can extend to one of them.

If N₀>K, then −1<A<0. The denominator is positive for t≥t₀ but vanishes at t_*=t₀+ln(−A)/r<t₀. The maximal connected mathematical interval containing t₀ is (t_*,∞), and N decreases toward K forward in time. A future prediction can be regular even though backward continuation fails after finite time.

For a negative initial value, which has no ordinary population interpretation, A<−1 and t_*=t₀+ln(−A)/r>t₀. The initial branch exists on (−∞,t_*) and becomes unbounded negatively as t approaches t_* from below. The equation is smooth at all finite states, but that does not prevent finite-time blow-up. Physical population modelling normally restricts to N≥0; the mathematical negative case clarifies the formula’s interval restrictions.

Derive the integrating factor from the product rule

Consider y′+p(t)y=q(t) on an interval I where p and q are continuous and t₀∈I. If the original equation is a(t)y′+b(t)y=c(t), first choose an interval where a(t)≠0 and set p=b/a, q=c/a. A zero of a requires its own original-equation analysis; it is not automatically included after division.

Define the normalized integrating factor μ(t)=exp(∫_(t₀)^t p(s) ds). Then μ(t₀)=1, μ>0 and μ′=pμ. Multiplication gives μy′+μpy=(μy)′=μq. Integration from t₀ to t yields μ(t)y(t)−y₀=∫_(t₀)^t μ(s)q(s) ds. Division by the positive μ is legitimate everywhere on I.

Therefore y(t)=μ(t)^(−1)[y₀+∫_(t₀)^t μ(s)q(s) ds]. This formula gives an actual differentiable solution across I and verifies the initial value. It also proves uniqueness there: the difference of two solutions has (μ(y₁−y₂))′=0 and initial value zero. Continuous linear coefficients on an interval support this whole-interval conclusion; it is stronger than a merely local rectangle guarantee.

Any nonzero constant multiple of μ works equally well in an indefinite derivation. The final solution still needs one independent integration constant or its initial-value replacement; it must not be discarded along with the arbitrary scaling of μ. When units are present, p has units inverse time, so ∫p dt is dimensionless, while q has units of y per time.

A linear solution separates retained data from accumulated input

The definite-integral solution can be written y(t)=G(t,t₀)y₀+∫_(t₀)^t G(t,s)q(s) ds, where G(t,s)=exp(−∫_s^t p(u) du). The first term transports the initial state through the homogeneous equation y′+py=0. For t≥t₀, the integral accumulates the forcing q over earlier times, weighted by the response between s and t. This identity follows by expanding the ratio μ(s)/μ(t) in the previous formula.

For constant p=k, the weight is e^(−k(t−s)). If k>0, earlier input is attenuated more strongly as its age increases. The homogeneous response decays forward in time. If k<0, it grows instead, and if k=0 the formula reduces to direct integration. The sign and units of k determine the interpretation; a linear equation is not necessarily a stable decay model.

If y_p is one particular solution and y_h is any homogeneous solution, y_p+y_h solves the same forced equation. The difference of two solutions to the same forced equation is homogeneous. Their sum generally doubles the forcing and is not another solution to the original forced equation. Superposition must be applied to the operator and the right sides actually present.

For y′+2y=6 with initial value y₀ at t₀, the equilibrium particular solution is 3 and the homogeneous response is C e^(−2(t−t₀)). Thus y=3+(y₀−3)e^(−2(t−t₀)). This familiar form is now derived from a general product-rule method. It is not necessary to guess that every forcing is constant: the integral formula remains available for continuous varying q.

Solve across a finite forcing switch with a clear solution notion

A forcing can change at a known time. For y′+ky=q(t) with a finite jump in q, solve separately on the open intervals to either side of the switch. In a model with no impulse and a continuous stored state, match the value of y at the switching time and use it as the next interval’s initial value. Continuity of the state is a modelling and solution condition here; it is not permission to claim the derivative is also continuous.

For example, let k=1 and q=0 for t<1, q=2 for t>1, with y(0)=0. Before the switch y=0. Matching y(1)=0 gives y=2(1−e^(−(t−1))) for t>1. Its left derivative at 1 is 0 and its right derivative is 2, so it is not a classical differentiable solution at the switch. It is continuous and piecewise classical, satisfying the integral balance across the switch.

Assigning any single finite value to q at the jump does not change its integral, but it cannot make these unequal one-sided derivatives equal. A globally classical solution claim therefore requires more than matching values. One may use a continuous piecewise C¹ solution with its integral equation in this finite-switch setting; more general weak formulations and impulses need their own definitions.

Do not introduce a jump in stored mass, charge or another state merely because its finite input rate jumps. A state jump would require a different model, such as an ideal impulse. Conversely, forcing that changes a coefficient rather than only q must be handled with the corresponding interval equations. Record every switching time and the continuity or jump conditions being imposed.

An exact differential relation preserves a potential

Write a first-order differential relation as M(x,y) dx+N(x,y) dy=0. Along a differentiable function y(x), this means M(x,y)+N(x,y)y′=0. If a C¹ potential H satisfies H_x=M and H_y=N on the domain, the chain rule gives dH(x,y(x))/dx=0. Thus each solution graph lies on a level curve H=C. This connects first-order equations to the potential and path-independence ideas of lessons 134–136.

With continuously differentiable M and N, the condition M_y=N_x is necessary for a C² potential. On an open rectangle it is also sufficient. To see this, fix (x₀,y₀) and set H(x,y)=∫_(x₀)^x M(s,y₀) ds+∫_(y₀)^y N(x,v) dv. Differentiation gives H_y=N and H_x=M(x,y₀)+∫_(y₀)^y N_x(x,v) dv=M(x,y), using N_x=M_y and the fundamental theorem.

This rectangle construction supplies a local potential near any point where M,N are C¹ and M_y=N_x on a neighbourhood. A global potential on a larger region requires suitable domain conditions, such as a simply connected open domain. Equality of cross-partials on a punctured plane does not by itself provide a global potential. The vortex field studied in lessons 134–136 is the standard warning, with a nonzero integral around its hole.

To find H in practice, integrate M with respect to x and retain an unknown function A(y), then differentiate in y and match N. The resulting H=C is an implicit relation. If N=H_y≠0 at an initial point, the implicit function theorem supplies a local differentiable graph y(x). At a point where that condition fails, inspect the original relation; a level curve may have a vertical tangent, multiple branches or a degenerate point.

A level curve is not automatically one global function

For 2x dx+2y dy=0, take H=x²+y². Through (0,2), the implicit relation is x²+y²=4. Near the initial point the correct graph is y=√(4−x²), with −2<x<2. It satisfies y′=−x/y and the original relation 2x+2yy′=0 throughout that interval. The lower semicircle is a different local branch, not the same initial-value solution.

At the endpoints (±2,0), the original relation would require 2x+0·y′=0, impossible for a finite derivative there. The upper semicircle therefore does not become a classical graph solution through either endpoint merely because the implicit circle continues as a geometric curve. Parametrised curves and functions of x answer different questions.

Conversely, an explicit formula can seem singular because it was derived by dividing a factor that the original equation allowed to vanish. Check those points directly. A removable algebraic denominator in one representation and a genuine failure of a classical solution are different possibilities. The original equation determines which is present.

If an implicit integral is not expressible in elementary functions, retain it with its branch and interval conditions. Local inversion or controlled numerical evaluation can still answer questions. Solving for y is an optional final convenience; declaring all implicit relations unsolved loses valid exact mathematical information.

A nonlinear exactness test can also reveal an integrating factor

An equation that is not exact may become exact after multiplication by a nonvanishing function μ. The exactness condition is (μM)_y=(μN)_x. If μ depends only on x, this expands to μM_y=μ′N+μN_x. On a region where N≠0, a sufficient recognisable case is (M_y−N_x)/N=A(x), giving μ(x)=exp(∫A(x) dx). Verify the multiplied cross-partials directly after choosing the factor.

If instead μ depends only on y, the expansion gives μ′M+μM_y=μN_x. Where M≠0, if (N_x−M_y)/M=B(y), choose μ(y)=exp(∫B(y) dy). The two tests have different denominators and signs. They are useful sufficient tests, not an exhaustive catalogue of integrating factors for every first-order relation.

For 2y dx+x dy=0 on x>0, M_y−N_x=1 and N=x, so A=1/x. Taking μ=x yields 2xy dx+x²dy=0, the exact differential of x²y. The original initial value y(1)=2 therefore gives y=2/x² on x>0. The same equation is also linear after division by x; the two methods agree because they are consistent rewritings.

Multiplication by a factor that is zero or undefined somewhere changes equivalence there. The exponential construction is nonzero only where its defining integral is meaningful. If a simplified factor such as y is used on a selected half-plane, analyse y=0 separately in the original equation. Likewise, a ratio test obtained by dividing N or M is initially justified only where that denominator is nonzero, even if a final factor later extends.

Bernoulli substitution linearises a selected nonzero branch

A Bernoulli equation has the form y′+p(x)y=q(x)yⁿ. Take p and q continuous on the interval being studied. The cases n=0 and n=1 are already linear and should be handled directly. For n≠0,1, select a nonzero real branch where the needed powers are defined and differentiable. For arbitrary real n, the positive branch y>0 is a convenient sufficient setting; other branches require their own real-power domain checks.

Divide by yⁿ and put v=y^(1−n). The chain rule gives v′=(1−n)y^(−n)y′, so the transformed equation is v′+(1−n)p(x)v=(1−n)q(x). This is linear in v. Solve it by an integrating factor, then invert the substitution only on a branch consistent with the original initial value and real-power domain.

For y′+y=2y² with y(0)=1/4, take v=1/y. Then v′−v=−2 and v(0)=4, giving v=2+2e^x and y=1/(2+2e^x). Direct substitution verifies the answer for all real x. The original equilibrium y=0 was excluded by taking a reciprocal and must be listed separately when describing the equation’s solutions.

A transformed solution need not stay in the range of the chosen power map. If it reaches zero or leaves that range, inspect the inverse and the original equation before extending. In the reciprocal example a zero of v would correspond to a pole in y. For other powers the behaviour can differ. If n<0, the original right side may itself be undefined at y=0, so zero is not automatically an equilibrium to append.

A known particular solution can reduce a Riccati equation

A Riccati equation is y′=a(x)y²+b(x)y+c(x); take a,b,c continuous on the chosen interval. There is no claim here that every such equation has an elementary solution. If one particular solution y_p is already known on an interval, however, the substitution y=y_p+1/v can reduce the remaining noncoincident branches to a linear equation wherever v is finite and nonzero.

Differentiate the substitution: y′=y_p′−v′/v². Expanding the quadratic right side and using y_p′=a y_p²+b y_p+c leaves −v′/v²=(2a y_p+b)/v+a/v². Multiplication by v² gives v′+(2a y_p+b)v=−a. Solve this linear equation, then reconstruct y on intervals where v≠0. The known y_p is retained separately because it has no finite v representation.

For y′=y²−1, choose y_p=1. The transformed equation is v′+2v=−1. With y(0)=0, v(0)=−1 and v=−(1+e^(−2x))/2. Hence y=1−2/(1+e^(−2x))=−tanh x, valid for all real x. Its derivative is −sech²x=y²−1. Both constant solutions y=1 and y=−1 can also be checked directly in the original equation.

This reduction is useful only when a particular solution is genuinely supplied or found. Guessing an incorrect y_p creates an incorrect linear equation even if the later integrating-factor algebra is flawless. The coefficients, chosen particular solution, reciprocal restrictions and initial conversion are all part of the method’s verification. No general Riccati solvability promise follows from this special reduction.

A ratio equation uses a different meaning of homogeneous

Some first-order equations have the form y′=F(y/x), on a domain with x≠0. They are traditionally called homogeneous first-order equations in the ratio sense. This terminology differs from a linear homogeneous equation with zero forcing. The ratio equation can be nonlinear, as y′=1+(y/x)² is.

Set v=y/x, so y=xv and y′=v+xv′. The equation becomes xv′=F(v)−v. On a branch where F(v)−v≠0, separation gives ∫dv/[F(v)−v]=ln|x|+C. Use an interval wholly on x>0 or x<0, as selected by the initial point. A formula derived using ln|x| does not justify passage through x=0.

Before dividing by F(v)−v, solve F(c)=c. Each such constant ratio gives the straight-line solution y=cx on a suitable original domain. For y′=y/x+(y/x)²−1, these ratios are c=±1. On other branches, xv′=v²−1 and the partial-fraction integral is (1/2)ln|(v−1)/(v+1)|=ln|x|+C.

For that example with y(1)=0, the initial ratio v(1)=0 gives v=(1−x²)/(1+x²), so y=x(1−x²)/(1+x²) on x>0. Although this rational expression itself is finite at zero, the original field contains y/x and is not defined there. The initial-value problem’s field domain therefore still blocks an unqualified continuation through zero.

Close the calculation in the original equation

A solution record should show the recognised structure, legitimate transformations, any divided or inverted factors, the integrated relation and the converted initial data. Differentiate the final candidate and substitute into the original equation. This final check can reveal a lost sign, wrong integrating-factor exponent or false particular solution that a transformed residual alone would miss.

Then give a connected interval containing the initial point. Its endpoints may come from original coefficient singularities, zeros of a reconstructed denominator, real-root limits, or loss of a differentiable graph. Do not copy the domain of an intermediate expression as though it automatically were the maximal solution interval. A sufficient working interval and a proved maximal interval should be labelled differently.

Retain excluded equilibria and known particular solutions as appropriate. Where nonuniqueness is possible, do not claim that one separated branch lists every solution. Where coefficients are continuous and a linear initial-value formula has been derived across an interval, its uniqueness follows from the product-rule argument. Match the strength of the conclusion to the argument actually available.

The explorer should display the original relation, selected branch, transformed variable where used, exact initial-data check and any inadmissible operation. Comparing direct residuals at several points helps test the displayed implementation but cannot prove an identity for every real input. Its diagrams should mark poles, switching times and vertical-tangent endpoints instead of connecting unrelated curve pieces.

Worked examples

Try it yourself

1. Is y′=(x²+1)(y−4) separable? Which constant solution must be recorded before dividing?

   Show the answer and reasoning

   Yes; y=4 is the equilibrium.

   The right side is a product of a function of x and a function of y. Dividing by y−4 removes its zero case.

2. If y′=a(x)b(y) and G′=1/b on a nonzero branch, what is dG(y(x))/dx?

   Show the answer and reasoning

   It is a(x).

   The chain rule gives G′(y)y′=y′/b(y)=a(x). The conclusion requires b(y)≠0 on the branch.

3. For y′=y(1−y), y(0)=1, should one divide by y(1−y) at the initial point?

   Show the answer and reasoning

   No; the original constant solution y=1 already satisfies the problem.

   The divided expression is undefined there. Smoothness and local uniqueness support retaining the equilibrium without introducing arbitrary departing branches.

4. From ln|y|=x+C and a negative initial value, may one choose y=e^C e^x without a minus sign?

   Show the answer and reasoning

   No; the initial sign must be restored.

   e^C is positive. The negative branch is y=−e^C e^x; an absolute value does not remove initial sign information.

5. For N′=2N(1−N/6), N(0)=12, where is the pole of the initial branch?

   Show the answer and reasoning

   At t=−(ln2)/2, with initial interval (−(ln2)/2,∞).

   The denominator 1−e^(−2t)/2 vanishes when e^(−2t)=2. The pole is backward from the initial time.

6. For y′+p(x)y=q(x), what normalized integrating factor has value 1 at x₀?

   Show the answer and reasoning

   μ(x)=exp(∫_(x₀)^x p(s) ds).

   Its derivative is pμ, making (μy)′=μq. Its positivity permits division throughout the continuous-coefficient interval.

7. If L[y₁]=q and L[y₂]=q for one linear differential operator, what are L[y₁−y₂] and L[y₁+y₂]?

   Show the answer and reasoning

   They are 0 and 2q, respectively.

   Linearity preserves the right sides under the same subtraction or addition. Two forced solutions cannot usually be added to get another solution with unchanged forcing.

8. Why is division of x y′+y=2x by x initially restricted to x≠0?

   Show the answer and reasoning

   The divided field contains 1/x and is undefined at zero.

   An original-equation solution may sometimes extend through zero, but that extension needs separate verification. Normal-form algebra alone cannot authorize it.

9. A continuous piecewise solution has left derivative 0 and right derivative 2 at a forcing switch. Is it a classical differentiable solution at that point?

   Show the answer and reasoning

   No.

   The unequal one-sided derivatives rule out differentiability there. A piecewise classical solution satisfying an integral balance is a different, explicitly stated notion.

10. For M=3x²+2y and N=2x+4y, construct a potential H.

    Show the answer and reasoning

    H=x³+2xy+2y², up to an additive constant.

    H_x=3x²+2y=M and H_y=2x+4y=N. Equality M_y=N_x=2 is consistent with exactness on the plane.

11. Does M_y=N_x on the whole punctured plane automatically guarantee one global potential?

    Show the answer and reasoning

    No.

    The domain can carry nonzero loop integrals around its hole. The local rectangle result is not an unconditional global theorem.

12. For H(x,y)=C with H_x=M and H_y=N, what convenient condition at an initial point gives a local differentiable graph y(x)?

    Show the answer and reasoning

    N=H_y≠0 there, with the required continuity of the derivatives.

    The implicit function theorem then supplies the graph. A level curve without this check can have a vertical tangent or other obstruction.

13. For an integrating factor μ(x), which ratio must depend only on x where its denominator is nonzero?

    Show the answer and reasoning

    (M_y−N_x)/N.

    Expanding (μM)_y=(μN)_x gives μ′/μ=(M_y−N_x)/N. The y-only test instead uses (N_x−M_y)/M.

14. For y′+p y=q yⁿ on a suitable nonzero real branch with n≠0,1, state the substitution and transformed linear equation.

    Show the answer and reasoning

    v=y^(1−n), with v′+(1−n)p v=(1−n)q.

    Multiply the divided equation by 1−n and use v′=(1−n)y^(−n)y′. Real powers and inverse branches retain their domain restrictions.

15. In a reciprocal substitution y=1/v, may a transformed solution be inverted through a zero of v as a finite classical y?

    Show the answer and reasoning

    No.

    The reciprocal is undefined at that point. The original initial branch and any continuation need separate analysis, and a pole often results.

16. For y′=a y²+b y+c with a known solution y_p, what equation follows from y=y_p+1/v?

    Show the answer and reasoning

    v′+(2a y_p+b)v=−a, where the substitution is valid.

    The y_p terms cancel because it solves the original equation. Multiplication by v² then gives a linear equation; y_p itself remains separately retained.

17. For y′=F(y/x), which constant ratios are lost by dividing the transformed equation by F(v)−v?

    Show the answer and reasoning

    The roots F(c)=c, giving y=cx on the original admissible domain.

    The substitution gives xv′=F(v)−v. Constant v=c solutions are precisely the zero cases of the divided factor.

18. Why must a final candidate be substituted into the original equation, rather than only a divided or transformed equation?

    Show the answer and reasoning

    Transformations can exclude states, change domains or introduce branch restrictions.

    The original residual, initial data and connected interval determine the requested solution. Correct algebra in a restricted transform alone may lose or misclassify valid cases.

What to carry forward

• Recognise the equation’s structure and domain before choosing a first-order method.
• Separation follows from the chain rule on a branch where the divided state factor is nonzero.
• Record equilibria before dividing, and use uniqueness hypotheses when discussing branch crossings.
• Logarithmic absolute values preserve branch information; restore the initial sign explicitly.
• A logistic formula has equilibrium and denominator cases that determine its initial branch’s interval.
• A normalized integrating factor turns a continuous linear equation into a product derivative.
• The definite-integral linear formula keeps the initial datum and accumulated forcing visible.
• The difference of solutions to the same linear forced equation is homogeneous; their sum changes the forcing.
• Finite forcing switches can give a continuous piecewise classical solution whose derivative jumps.
• An exact differential relation preserves a potential, with local and global domain conditions kept distinct.
• An implicit level curve must be restricted to an actual differentiable function branch.
• One-variable integrating-factor tests are sufficient recognisable cases, with signs and denominators that matter.
• Bernoulli substitution linearises a selected nonzero real branch and can omit zero solutions.
• A known Riccati particular solution reduces other admissible branches to a linear equation, while remaining separately retained.
• Ratio substitution requires x≠0 and separately retains constant-ratio straight lines.
• Verify every final candidate, initial datum and interval in the original equation; a transformed formula alone is insufficient.