Separable Initial-Value Problems With Multiple Branches

Why an Implicit Relation Can Give More Than One Candidate

A separable differential equation may lead to an implicit relation that has several explicit forms. Solving that relation can introduce a choice of sign, and each sign can describe a different curve. The differential equation alone may permit both curves, but an initial value specifies a point the solution must pass through. That point determines which branch is relevant.

Consider the equation $y'=\frac{2x}{y}$. Its right-hand side is defined only when $y\ne 0$. Separating variables and integrating leads to a relation involving $y^2$. Because squaring does not retain the sign of $y$, solving that relation for $y$ gives both a positive and a negative candidate. We will determine exactly how an initial value selects between them, and why the interval containing the initial point matters.

Definition: An initial-value problem (IVP) consists of a differential equation together with an initial condition $y(x_0)=y_0$. A solution must satisfy the equation on an interval containing $x_0$, and it must have the specified value $y_0$ at $x_0$.

From the Differential Equation to the Implicit Relation

For $y\ne 0$, multiplying $y'=\frac{2x}{y}$ by $y$ gives $yy'=2x$. Since $\frac{d}{dx}(y^2)=2yy'$, integration gives a relation between $y^2$ and $x^2$:

yy'=2x \quad\Longrightarrow\quad \frac{1}{2}y^2=x^2+C \quad\Longrightarrow\quad y^2=2x^2+C_1.

Here $C$ and $C_1$ are constants; the final equation uses $C_1=2C$. An initial condition $y(x_0)=y_0$ determines the constant by substitution:

$$ C_1=y_0^2-2x_0^2. $$

The relation $y^2=2x^2+C_1$ is implicit: it relates $x$ and $y$ without expressing $y$ as a single formula. Solving for $y$ produces two possible signs wherever the square root is positive.

Key point: The equation $y^2=2x^2+C_1$ describes possible points on solution curves, but it does not by itself say which sign of $y$ applies. An initial value with $y_0\ne 0$ selects the sign by requiring the curve to pass through $(x_0,y_0)$.

Why the Initial Value Selects a Branch

There is a useful justification that also shows why the implicit relation captures every solution on an interval where the differential equation is defined. Suppose $y$ is a differentiable solution on an interval $I$, with $y(x)\ne 0$ throughout $I$. Define $F(x)=y(x)^2-2x^2$. Differentiating and using the differential equation gives:

F'(x)=2y(x)y'(x)-4x =2y(x)\left(\frac{2x}{y(x)}\right)-4x =0.

For any two points in $I$, the mean value theorem applied between them shows that a differentiable function with derivative zero throughout the interval has the same value at both points. Therefore $F$ is constant on $I$, so $y^2-2x^2=C_1$ there. This proves that every solution must satisfy the implicit relation.

Now suppose the initial value has $y_0\ne 0$. A continuous function that never equals zero on an interval cannot change sign there: if it changed from positive to negative, the intermediate value theorem would force it to equal zero at some point. Thus the sign throughout the solution interval must match the sign at $x_0$. Solving the implicit relation with that sign gives the selected branch.

Theorem: For $y'=\frac{2x}{y}$ with $y_0\ne 0$, the initial condition $y(x_0)=y_0$ determines $C_1=y_0^2-2x_0^2$ and selects the branch $$ y(x)=\operatorname{sgn}(y_0)\sqrt{2x^2+C_1} $$ on an interval containing $x_0$ where $2x^2+C_1>0$. Here $\operatorname{sgn}(y_0)=1$ if $y_0>0$, and $\operatorname{sgn}(y_0)=-1$ if $y_0<0$.

The condition $2x^2+C_1>0$ ensures that the square root is real and that $y\ne 0$, as required for the differential equation. The branch must be considered on the part of this set lying in an interval that contains the initial point. If that set has separate pieces, a solution through one initial point cannot be taken from a piece that does not contain it.

Worked Examples: Selecting and Checking a Branch

Worked Example: A Positive Initial Value

Solve $y'=\frac{2x}{y}$ with $y(1)=2$, and check the selected solution.

First calculate the constant from the implicit relation:

$$ C_1=y_0^2-2x_0^2 =2^2-2(1)^2 =4-2 =2. $$

Thus $y^2=2x^2+2$. At $x=1$, the required value is positive, so select the positive branch:

y(x)=\sqrt{2x^2+2}.

The radicand is positive for every real $x$, so this formula has no zero and is defined on all of $\mathbb R$. At the initial point, $y(1)=\sqrt{2+2}=2$, as required.

To check the differential equation, differentiate:

y'(x)=\frac{1}{2}(2x^2+2)^{-1/2}(4x) =\frac{2x}{\sqrt{2x^2+2}} =\frac{2x}{y(x)}.

Both the equation and initial condition hold. The negative branch would give $y(1)=-2$, so it does not solve this IVP.

Worked Example: A Negative Initial Value

Solve $y'=\frac{2x}{y}$ with $y(0)=-3$, and verify the answer.

The constant is:

$$ C_1=y_0^2-2x_0^2 =(-3)^2-2(0)^2 =9-0 =9. $$

The implicit relation is $y^2=2x^2+9$. Since the initial value is negative, the solution must use the negative branch:

y(x)=-\sqrt{2x^2+9}.

At $x=0$, this gives $y(0)=-\sqrt{9}=-3$. To check the differential equation, differentiate the chosen branch:

y'(x) =-\frac{1}{2}(2x^2+9)^{-1/2}(4x) =-\frac{2x}{\sqrt{2x^2+9}}.

Since $y(x)=-\sqrt{2x^2+9}$, the right-hand side of the equation is also:

\frac{2x}{y(x)} =\frac{2x}{-\sqrt{2x^2+9}} =-\frac{2x}{\sqrt{2x^2+9}} =y'(x).

The radicand is positive for every real $x$, so the formula stays away from $y=0$. It satisfies both the equation and the initial condition on $\mathbb R$.

Worked Example: A Branch on One of Two Separate Intervals

Solve $y'=\frac{2x}{y}$ with $y(2)=1$. Determine which part of the implicit relation gives a solution on an interval containing $x=2$, and check it.

Calculate the constant:

$$ C_1=y_0^2-2x_0^2 =1^2-2(2)^2 =1-8 =-7. $$

The implicit relation is $y^2=2x^2-7$, so its explicit forms require $2x^2-7>0$. Solving this inequality gives $x^2>\frac{7}{2}$, or $x<-\sqrt{\frac{7}{2}}$ or $x>\sqrt{\frac{7}{2}}$. The initial point $x=2$ lies in the second interval, because $2>\sqrt{\frac{7}{2}}$. Also, $y(2)=1$ is positive, so the selected branch is:

y(x)=\sqrt{2x^2-7}, \qquad x>\sqrt{\frac{7}{2}}.

At $x=2$, $y(2)=\sqrt{8-7}=1$. Differentiating on this interval gives:

y'(x)=\frac{1}{2}(2x^2-7)^{-1/2}(4x) =\frac{2x}{\sqrt{2x^2-7}} =\frac{2x}{y(x)}.

The positive branch on $x<-\sqrt{\frac{7}{2}}$ does not contain the initial point $x=2$. The negative branch on the interval containing $2$ gives $y(2)=-1$, not $1$. The selected formula therefore satisfies both requirements on the interval containing the specified data.

A Common Mistake: Keeping Both Signs After Applying the Data

When solving $y^2=2x^2+2$, it is correct to write two algebraic possibilities before using an initial condition:

y=\sqrt{2x^2+2} \qquad\text{or}\qquad y=-\sqrt{2x^2+2}.

A common error is to present both as solutions to the IVP $y'=\frac{2x}{y}$, $y(1)=2$. Substituting the initial point shows why that is wrong:

\sqrt{2(1)^2+2}=2, \qquad -\sqrt{2(1)^2+2}=-2.

Only the positive branch has the required value. The correction is to use the initial value after solving the implicit relation, not to carry both signs forward as though both met the same initial condition. Both signs may solve the differential equation on suitable intervals; they do not both solve this particular IVP.

When the Initial Value Is Not Admissible

The differential equation $y'=\frac{2x}{y}$ is not defined at any point where $y=0$. Therefore an initial condition with $y_0=0$ cannot be satisfied by a classical solution of this equation at the initial point. For instance, $y(0)=0$ would make the expression $\frac{2x}{y}$ undefined at $x=0$. An implicit curve passing through $(0,0)$ does not remove that problem: the original equation itself must hold wherever the solution is defined, including at the initial point.

There is a related issue even for nonzero initial values: a formula cannot be extended through a point where its value becomes zero, because the differential equation is undefined there. For this reason, selecting a sign and identifying an interval containing the initial point are both part of specifying the solution. The initial value chooses a branch locally on its interval; it does not make every part of the implicit curve a solution through that point.

Branch Selection as a Repeatable Method

For an IVP that produces a squared relation, use a careful sequence rather than choosing a sign by inspection alone:

Find the implicit relation.
Separate and integrate on a region where the algebraic steps are valid, retaining any restrictions from the original equation.

Use the initial data to determine the constant.
Substitute $x_0$ and $y_0$ into the relation and compute the constant explicitly.

Resolve the sign using $y_0$.
If $y_0>0$, select the positive square root; if $y_0<0$, select the negative square root.

Choose a valid interval containing $x_0$.
Require the radicand to be positive so the selected branch is real and nonzero for this equation.

Verify the result.
Check the initial value and differentiate the explicit formula to confirm that it satisfies the differential equation.

This method separates three questions that are easy to confuse: what implicit relation the differential equation gives, which sign matches the initial value, and where the resulting formula satisfies the original equation. Checking all three prevents an algebraically plausible branch from being mistaken for the solution to the IVP.

Check Your Understanding

For each question, distinguish the implicit relation, the selected sign, and the interval needed for a valid solution.

For $y'=\frac{2x}{y}$ with $y(1)=-2$, what is the constant in $y^2=2x^2+C_1$, and which branch does the initial condition select?
Why can a continuous solution that never equals zero not change sign on its interval?
For $y'=\frac{2x}{y}$ with $y(2)=1$, why is the branch on $x<-\sqrt{\frac{7}{2}}$ not the solution through the initial point?
What is the problem with prescribing $y(0)=0$ for the equation $y'=\frac{2x}{y}$?
After finding an explicit branch, what two checks confirm that it solves the IVP?

Separable Initial-Value Problems With Multiple Branches

What You'll Learn

Why an Implicit Relation Can Give More Than One Candidate

From the Differential Equation to the Implicit Relation

Why the Initial Value Selects a Branch