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Calculus BC · Unit 7 · 7.7
AP Calculus · Differential Equations
Particular Solutions Using Initial Conditions
One point selects one curve — and also decides the branch and the domain.
key idea$\text{One initial condition selects \emph{one} curve from the whole family}$
The procedure
A general solution is a family; an initial condition is the address of one member. Separate and integrate as usual, then substitute the given point to pin down $C$. The one genuinely new concern is the domain: a particular solution lives on a single interval, not everywhere.
1
Separate and integrate
As for any separable equation, ending with a constant $C$ on the $x$-side.
in equation $+$ point out a relation with $C$
2
Substitute the initial condition immediately
Plug the point in while the relation is still implicit. Solving for $C$ before exponentiating avoids the absolute-value branch problem entirely.
out a numeric $C$
3
Resolve the sign or branch
The initial point tells you which side of a $\pm$ or an absolute value you are on. Choose that branch and discard the other.
out one function, not two
4
Solve for $y$ explicitly
Exponentiate or take the appropriate root, now that $C$ and the branch are both fixed.
out $y=$ something in $x$
5
State the domain
The solution runs only as far as the largest interval containing the initial $x$ on which it stays defined and differentiable.
out the interval of validity
| Initial condition | General solution | Fixing $C$ | Particular solution |
|---|
| $y(0)=5$ | $y=Ce^{2x}$ | $5=Ce^{0}$ | $y=5e^{2x}$ |
| $y(0)=-3$ | $y=Ce^{2x}$ | $-3=C$ | $y=-3e^{2x}$ — note $C\lt 0$ is fine |
| $y(1)=2$ | $y^{2}=x^{2}+C$ | $4=1+C$ | $y=+\sqrt{x^{2}+3}$, the positive branch |
| $y(0)=1$ | $y=\dfrac{1}{C-x}$ | $1=\dfrac1C$ | $y=\dfrac{1}{1-x}$ on $x\lt 1$ only |
Row 3 shows the branch decision and row 4 the domain restriction — the two places where a particular solution demands more care than a general one. The initial point settles both.
what the initial condition decides
- Substitute the point early, while the relation is still implicit. It is easier and safer than at the end.
- The sign of the initial $y$-value chooses the branch of a $\pm$ or an absolute value.
- A particular solution's domain is the largest interval around $x_0$ on which the formula stays defined.
- The solution curve can never cross an equilibrium solution, so equilibria bound the domain too.
Worked examples
EX 1
Substitute early
$\dfrac{dy}{dx}=2xy$, $y(0)=3$Solve $\dfrac{dy}{dx}=2xy$ with $y(0)=3$.
1
Separate and integrate
$\dfrac{dy}{y}=2x\,dx$ gives $\ln\lvert y\rvert=x^{2}+C$.
2
Substitute the point now
$x=0$, $y=3$: $\ln3=0+C$, so $C=\ln3$.
3
Rewrite
$\ln\lvert y\rvert=x^{2}+\ln3$.
4
Exponentiate
$\lvert y\rvert=3e^{x^{2}}$. Since $y(0)=3\gt 0$, the solution stays positive: $y=3e^{x^{2}}$.
5
Note the pay-off
Substituting before exponentiating turned the absolute value into a one-line sign check.
EX 2
Choosing a branch
$\dfrac{dy}{dx}=\dfrac{x}{y}$, $y(0)=-2$Solve $\dfrac{dy}{dx}=\dfrac{x}{y}$ with $y(0)=-2$.
1
Separate and integrate
$y\,dy=x\,dx$ gives $\dfrac{y^{2}}{2}=\dfrac{x^{2}}{2}+C$, so $y^{2}=x^{2}+K$.
2
Fix the constant
$(-2)^{2}=0+K$, so $K=4$ and $y^{2}=x^{2}+4$.
3
Choose the branch
$y=\pm\sqrt{x^{2}+4}$, and $y(0)=-2$ is negative.
4
Write the solution
$y=-\sqrt{x^{2}+4}$.
5
Confirm the domain
$x^{2}+4\gt 0$ always, so this one is valid for all real $x$ — not every problem restricts.
EX 3
A finite domain
the solution blows upSolve $\dfrac{dy}{dx}=y^{2}$ with $y(0)=1$ and state where the solution is valid.
1
Separate
$\dfrac{dy}{y^{2}}=dx$, so $\displaystyle\int y^{-2}dy=\int dx$.
2
Integrate
$-\dfrac1y=x+C$.
3
Fix the constant
$-\dfrac11=0+C$ gives $C=-1$, so $-\dfrac1y=x-1$.
4
Solve for $y$
$y=\dfrac{1}{1-x}$.
5
State the domain
The formula fails at $x=1$. The interval containing $x_0=0$ is $x\lt 1$, and $y\to\infty$ as $x\to1^{-}$ — a finite-time blow-up from a perfectly tame-looking equation.
EX 4
Blocked by an equilibrium
$\dfrac{dy}{dx}=y(y-4)$-style reasoningFor $\dfrac{dy}{dx}=2(y-5)$ with $y(0)=3$, find the particular solution and explain why $y$ never reaches $5$.
1
Separate and integrate
$\dfrac{dy}{y-5}=2\,dx$ gives $\ln\lvert y-5\rvert=2x+C$.
2
Substitute the point
$\ln\lvert3-5\rvert=\ln2=C$.
3
Resolve the absolute value
$y(0)=3\lt 5$, so $y-5$ is negative: $5-y=2e^{2x}$.
5
Explain the barrier
$y=5$ is itself a constant solution, and solutions cannot cross. Here $2e^{2x}\gt 0$ always, so $y$ stays strictly below $5$ — and in fact runs away downward.
Where it goes wrong
Substitute the point before exponentiating
Solving for $C$ inside the logarithm is one line. Solving for it after exponentiating drags the absolute value through the algebra.
A $\pm$ is a decision, not an answer
The initial value's sign picks one branch. Leaving both is an incomplete answer to a particular-solution question.
State the domain when the formula can fail
If the denominator vanishes or a radicand goes negative, the solution stops there. The valid interval is the one containing $x_0$.
Equilibrium solutions are walls
A particular solution cannot pass through a constant solution. That fact alone often answers ‘what happens in the long run’.