01 · A negative initial value
For y′ = 2xy and y(1) = −2, find y.
Hint
Use the same family.
Worked solution
y = −2e^(x² − 1).
Understand · explore · practise
Differential equations: initial conditions and domains
Find particular differential-equation solutions using initial data. Choose signs, retain equilibrium solutions, identify connected domains around poles and circle endpoints, and reject incompatible initial conditions.
Before you startSeparable differential equations, logarithms, implicit differentiation and graph domains.
01 / An initial condition selects a solution branch
Find the constant and the interval containing the given point.
Check equation + initial value + domain
A formula is not a complete particular solution if its chosen branch misses the initial point.
After integrating, substitute the supplied point to determine the constant. If a square root or inverse function appears, select the branch consistent with that point. Finally keep a connected interval on which the original differential equation and the solution are defined.
Keep the branch through the initial pointExplore
y = 1/(1 − x).
Connected solution interval through x = 0: x < 1.
At x = 0, y = 1.
Derivative = 1; original right-hand side = 1.
Dashed lines mark excluded finite interval boundaries. A formula can exist on another interval without making that disconnected branch part of this initial-value solution. Values outside the displayed y-range are clipped visually; the written domain remains authoritative.
02 / Substitute the point into the general family
Use the actual x-coordinate, not automatically zero.
Solve y′ = 2xy with y(1) = 3.Worked example
y = A e^(x²)
This is the general family, including A = 0.
3 = Ae, so A = 3/e
Substitute both coordinates.
y = 3e^(x² − 1)
The formula is defined for every real x.
Check y(1) = 3 and y′ = 2xy
Both requirements hold.
02 · A zero initial value
For y′ = 2xy and y(1) = 0, find the solution in this family.
Hint
The exponential is never zero.
Worked solution
A = 0, so y = 0.
03 / A pole limits the initial-value interval
Do not continue a solution through a vertical asymptote.
Solve y′ = y² with y(0) = 2.Worked example
For y ≠ 0, y = −1/(x + C)
Separate and integrate.
2 = −1/C, so C = −1/2
Apply the initial condition.
y = 2/(1 − 2x)
The denominator vanishes at x = 1/2.
The interval containing zero is x < 1/2
The disconnected branch x > 1/2 is not the continuation of this initial-value solution.
Watch: keep the branch through the initial point
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Another pole
For y′ = y² and y(0) = −1, give the solution and maximal interval containing zero.
Hint
The formula is −1/(1 + x).
Worked solution
y = −1/(1 + x), with x > −1.
04 · The equilibrium exception
What happens for y′ = y² and y(0) = 0?
Hint
Check the constant zero function directly.
Worked solution
y = 0 for all real x. It has no pole.
04 / Choose the sign from the initial value
Squaring an equation loses sign information.
Solve y′ = x/y with y(0) = −2.Worked example
y² = x² + C
Integrate y dy = x dx.
4 = C
Use the initial point.
y = −√(x² + 4)
Choose the negative branch to match y(0) = −2.
All real x are allowed
The radicand is positive and y never reaches zero.
05 · Positive branch
For the same differential equation with y(0) = 2, choose the branch.
Hint
The constant is still 4.
Worked solution
y = √(x² + 4), for all real x.
06 · Check a chosen branch
Differentiate y = −√(x² + 4) and compare with x/y.
Hint
Use the chain rule.
Worked solution
y′ = −x/√(x² + 4) = x/y.
05 / An implicit circle gives local function branches
Vertical endpoints are excluded from this derivative equation.
Solve y′ = −x/y with y(0) = 2.Worked example
y dy = −x dx → x² + y² = C
Integrate.
C = 4
The initial point is (0,2).
y = √(4 − x²)
The initial value selects the upper semicircle.
−2 < x < 2
At either endpoint y = 0 and the original quotient is undefined.
The full circle is an implicit curve, but it is not one differentiable function y(x) across its vertical endpoints. The endpoints can belong to the geometric circle while being excluded from the differential-equation solution.
07 · Lower semicircle
Use y(0) = −2 in y′ = −x/y.
Hint
Select the other sign.
Worked solution
y = −√(4 − x²), on −2 < x < 2.
08 · Can the endpoint be included?
May (2,0) be included in a solution of y′ = −x/y?
Hint
Evaluate the original denominator.
Worked solution
No. The derivative equation is undefined at y = 0.
06 / Absolute values record multiple possible branches
The initial point decides the sign of a nonzero expression.
Solve y′ = (y − 1)/(x + 2) with y(0) = 4.Worked example
For y ≠ 1: dy/(y − 1) = dx/(x + 2)
The original equation requires x ≠ −2.
ln|y − 1| = ln|x + 2| + C
Integrate on a connected interval.
y − 1 = A(x + 2)
A absorbs the branch sign and positive exponential constant.
3 = 2A, so A = 3/2
Use the initial condition.
y = 1 + (3/2)(x + 2), x > −2
The simplified line exists at −2, but the original equation does not.
09 · A removable-looking exclusion
Why is x = −2 excluded even when the simplified solution is a straight line?
Hint
Check the original equation, not only the final formula.
Worked solution
The right-hand side divides by x + 2, so it is undefined at −2.
07 / Check initial data before forcing an answer
Some points cannot satisfy the original equation.
Can (y + 2)y′ = 2x + 3 have a differentiable solution through (0,−2)?Worked example
Substitute x = 0 and y = −2
The left coefficient y + 2 becomes zero.
0 × y′ = 3
No finite derivative can satisfy this.
No differentiable solution passes through that point
Do not report an integrated relation as a valid solution there.
10 · An undefined initial quotient
Can y′ = x/y use the initial point (0,0)?
Hint
The original expression has denominator zero.
Worked solution
No. The original differential equation is undefined there, even though algebraically modified relations may pass through the point.
11 · An admissible zero
Why is y(0) = 0 allowed for y′ = y²?
Hint
There is no division by y in the original equation.
Worked solution
The right-hand side is defined and zero. The equilibrium y = 0 satisfies it.
08 / Report a connected interval containing the initial x
A list of excluded points is not always enough.
A proposed solution has poles at x = −2 and x = 3, and its initial x is 1.Worked example
The poles split the real line into three intervals
They are x < −2, −2 < x < 3 and x > 3.
The initial x = 1 lies between the poles
Choose the middle connected interval.
Use −2 < x < 3, subject to any further restrictions
Do not join disconnected pieces across either pole.
12 · Select the correct side
A solution has a pole at x = 4 and the initial point has x = 5. Which maximal side contains it?
Hint
Compare 5 with 4.
Worked solution
x > 4, if there are no other restrictions.
09 / Verify an implicit answer without solving explicitly
Differentiate the relation and compare on its valid domain.
Verify x² + y² = 9 for y′ = −x/y.Worked example
2x + 2yy′ = 0
Differentiate implicitly.
y′ = −x/y when y ≠ 0
Divide only where allowed.
Choose a branch and interval matching any initial point
The implicit relation alone does not identify which semicircle.
13 · Check an implicit power family
Differentiate y² = x³ + C.
Hint
Use the chain rule on y².
Worked solution
2yy′ = 3x², hence y′ = 3x²/(2y) where y ≠ 0.
10 / Check all three parts of the answer
A correct derivative can still accompany a wrong initial value.
Test the initial coordinates, verify the derivative equation, and inspect the original domain. If the formula is implicit, explain the branch when the initial condition selects one. If division discarded an equilibrium, check whether that equilibrium is the required solution.
14 · Wrong sign, correct implicit curve
A student gives y = √(4 − x²) for y(0) = −2. What failed?
Hint
Evaluate at zero.
Worked solution
The formula gives +2. It lies on the correct circle but selects the wrong branch.
15 · Formula beyond the pole
For y = 2/(1 − 2x), the formula has a value at x = 1. Does that extend the solution from y(0) = 2 to x = 1?
Hint
A pole lies between zero and one.
Worked solution
No. The interval containing the initial point ends at 1/2; a disconnected formula branch is not a continuation through the pole.
11 / Use the point, choose the branch, keep the domain
The original equation is the final authority.
- Check the supplied point is admissible.
- Find the integration constant using both coordinates.
- Choose the sign or branch that matches the initial value.
- Retain valid equilibrium solutions.
- Locate poles, zero denominators and root endpoints.
- Use a connected interval containing the initial x.
- Verify both the equation and the initial condition.
16 · Final initial-value problem
For y′ = y² with y(0) = 1/2, give y and its maximal interval containing zero.
Hint
Use y = y₀/(1 − y₀x).
Worked solution
y = 1/(2 − x), with x < 2.
Section 1 of 11 · An initial condition selects a solution branch