Separate variables, impose initial values and retain solution domains

Use the completed integration and verification foundations

Senior 5A–5B supplies integration rules and 6A supplies derivative identities, initial-value checks and domains. The companion lessons reuse existing y′=f(x) initial-value tasks and polynomial acceleration integration. Here the new step is solving a genuinely y-dependent equation

Recognise the separable structure

If y′=f(x)g(y), first check constant solutions from g(y)=0 in the original equation. On a branch with g(y)≠0, separate dy/g(y)=f(x)dx, integrate and combine the two additive constants into one. Then impose initial values and verify on a connected valid interval

Separation is justified by the chain rule

If G′(y)=1/g(y), then dG(y(x))/dx=y′/g(y)=f(x). Integrating gives G(y)=F(x)+C. The differential notation records this chain-rule argument; it does not permit dropping terms from a sum such as x+y.

Which step is valid for y′=3x(y-2) on a branch y≠2?

Exponentiation changes the role of the constant

For y′=k(y-h), separation gives ln|y-h|=kx+C on a nonzero branch. Thus y=h+D e^(kx), where D is a signed nonzero constant. The sign comes from the branch; e^C alone is positive. Check y=h separately before including D=0 in the final family

Work a signed exponential initial value

Solve y′=-2(y-3), y(0)=1. The family is y=3+D e^(-2x); the initial condition gives D=-2. Hence y=3-2e^(-2x), defined for all real x. The equilibrium y=3 is valid but does not meet this initial condition

For y′=-2(y-3), y(0)=1 and y=3+D e^(-2x), give D.

An x factor changes the exponent

For y′=2x(y+1), y(0)=2, separate dy/(y+1)=2x dx. Then ln|y+1|=x²+C, so y=-1+D e^(x²). The condition gives D=3. Differentiation recovers 2x(y+1); the all-real domain is valid

For y′=2x(y+1) and y=-1+D e^(Ax²), give A.

For y′=2x(y+1), y(0)=2 and y=-1+D e^(x²), give D.

A reciprocal solution can have a finite pole

For y′=y², y(0)=1, separate y^(-2)dy=dx. Integrating gives -1/y=x+C; the condition gives C=-1. Thus y=1/(1-x). Its maximal connected interval through 0 is x<1, although the same expression has a separate right-hand branch. The omitted equilibrium y=0 is also valid

Which branch belongs to the initial-value problem shown?

Give the pole of y=1/(1-x)

An implicit relation needs a branch and a strict domain

For y′=-x/(y-2), y(0)=4, separation gives (y-2)²=4-x². The initial value selects y=2+√(4-x²). The maximal interval is -2<x<2: at either endpoint the original denominator is zero. The lower square-root branch solves a different initial-value problem, y(0)=0.

For y′=-x/(y-2), y(0)=4 and (y-2)²=4-x², give the square of the maximal interval's radius

A circular x factor can be integrated before exponentiating

Solve y′=3cos(2x)y, y(0)=-2. On the nonzero branch, ln|y|=(3/2)sin(2x)+C, so y=-2e^((3/2)sin(2x)). At x=π/2 the sine is zero and y returns to -2. The solution remains negative and is defined for all real x; y=0 is the omitted equilibrium

Give the coefficient of sin(2x) in the exponent

An inverse circular primitive determines a tangent branch

For y′=2(1+y²), y(0)=1, integrate dy/(1+y²)=2dx to get arctan y=2x+π/4. Thus y=tan(2x+π/4). On the connected branch through 0, require -π/2<2x+π/4<π/2, giving -3π/8<x<π/8. The right-hand side has no real zero

For y=tan(2x+π/4) with y(0)=1, give the left endpoint of its maximal interval as a coefficient of π

For y=tan(2x+π/4) with y(0)=1, give the right endpoint of its maximal interval as a coefficient of π

Integrate a prescribed second derivative twice

For y″=f(x), integrate first to y′=F(x)+C1, then to y=G(x)+C1x+C2. Conditions on y′ and y determine the constants. Do not reuse one constant or drop the linear term. This is different from y″ depending on y, which is outside this simple rule

Work exponential forcing with two conditions

Solve y″=4e^(2x)+2, y(0)=3, y′(0)=-1. First y′=2e^(2x)+2x+C, so C=-3. Then y=e^(2x)+x²-3x+D, so D=2. Differentiate twice and check both initial values; the solution is defined for all real x.

For y″=4e^(2x)+2, y′(0)=-1 and y′=2e^(2x)+2x+C, give C.

For y=e^(2x)+x²-3x+D with y(0)=3, give D.

Higher powers can create two finite blow-up boundaries

For y′=xy³, y(0)=-2, separation gives -1/(2y²)=x²/2+C. The condition gives 1/y²=1/4-x², hence y=-1/√(1/4-x²). Its maximal interval is -1/2<x<1/2. This solution blows up at both endpoints; the original polynomial rate rule itself is defined for every finite (x,y)

For y′=xy³, y(0)=-2 and y=-1/√(B-x²), give B.

For y′=xy³, y(0)=-2 and y=s/√(1/4-x²), give s as +1 or -1.

Omitted equilibrium and forbidden denominator are different

For y′=xy³, division by y³ omits the valid solution y=0. For y′=x/y, the original equation is undefined at y=0, so it cannot be restored as an equilibrium. Return to the original equation whenever an algebraic step excludes a value

General solution does not mean arbitrary global switching

After separation, preserve differentiability, all initial conditions and a connected domain. An implicit relation may define several branches; joining them is not automatically a solution. Local uniqueness can help where its hypotheses hold, but it does not remove a finite pole

Derive and classify a separated family, ungraded

Solve y′=2x(y-1) for all real initial values y(0)=m. Derive the nonzero branches, recover the equilibrium and justify the domain and sign of y-1.

Model: retain the omitted constant solution

For y′=2x(y-1) and y(0)=m, separate on y≠1: ln|y-1|=x²+C, hence y=1+D e^(x²) with D≠0. Direct substitution shows y=1 also solves the original equation, so the complete initial-value family is y=1+(m-1)e^(x²). It is differentiable for every real x and has initial value m. Since the exponential is positive, y-1 has the fixed sign of m-1; m=1 gives the equilibrium. The smooth right-hand side supplies local uniqueness, so no other solution through that initial point can branch away while remaining finite

Solve a shifted reciprocal initial value, ungraded

Solve y′=-2y², y(1)=-1. State the maximal connected domain, verify the derivative and initial value, and explain the equilibrium lost during division

Model: locate the pole before choosing the interval

Separation gives -1/y=-2x+C; at x=1,y=-1 this yields C=3. Thus y=1/(2x-3), with pole x=3/2. The initial point lies to its left, so the maximal interval is (-∞,3/2). Differentiating gives y′=-2/(2x-3)²=-2y², and y(1)=-1. The separate y=0 solves the equation but not this initial condition. The right-hand branch is a different solution domain

Explain the implicit branches, ungraded

Solve y′=-x/(y-2), y(0)=0. Derive the implicit equation, choose the root sign and justify why endpoints cannot be included or connected to the other branch

Model: the original equation decides endpoint validity

Integrate (y-2)dy=-x dx to get (y-2)²+x²=C. The initial point gives C=4 and y-2<0, hence y=2-√(4-x²). Require -2<x<2; at ±2 the value y=2 makes the original rate undefined. The derivative x/√(4-x²) equals -x/(y-2) on that open interval and the initial value is 0. Switching to the upper branch away from the excluded endpoints would violate continuity; joining at an endpoint would violate the original equation

Solve prescribed polynomial and circular second derivatives, ungraded

Solve (i) y″=6x+4 with y(1)=2,y′(1)=-1; (ii) y″=-4sin(2x) with y(0)=1,y′(0)=0. Retain all constants and verify the results. The companion frozen acceleration task exercises the same componentwise principle

Model: two integrations leave a linear and a constant term

For (i), y″=6x+4 with y(1)=2 and y′(1)=-1: y′=3x²+4x+C1; its condition gives C1=-8. Then y=x³+2x²-8x+C2, and y(1)=2 gives C2=7. For (ii), y″=-4sin(2x) with y(0)=1 and y′(0)=0: y′=2cos(2x)+C1, so C1=-2. Then y=sin(2x)-2x+C2, and C2=1. Both functions are defined for all real x. Direct differentiation and both initial checks confirm each solution; the two constants represent different freedoms

Derive a contextual solution and check its limits, ungraded

An idealised 40 L well-mixed tank receives and loses 4 L/min, with inlet dye concentration 1.5 g/L and initial dye mass 20 g. Solve the mass equation, state the long-time behaviour and assumptions, and describe reproducible technology checks without claiming they prove the model

Model: solve the displacement from equilibrium

For a well-mixed 40 L tank with equal 4 L/min inlet and outlet flows, inlet dye concentration 1.5 g/L and initial dye mass 20 g, let M be dye mass in grams and t be time in minutes. The mass equation is M′=6-M/10=-(M-60)/10. Separating the non-equilibrium branch gives M=60+D e^(-t/10), and M(0)=20 gives D=-40. Thus M=60-40e^(-t/10) for physical times t≥0, approaching 60 g from below. Verify M′=4e^(-t/10)=6-M/10 and the initial mass. Equal flows, constant inlet concentration, perfect mixing and no reactions are assumptions. A documented plot or numerical solver can compare residuals, starting value and monotone approach on a stated time interval; refine tolerances and compare with data to assess assumptions. Full derivation and independent execution remain ungraded