How to Solve $2\sqrt{xy}\,\frac{dy}{dx}=1$

The equation

$$2\sqrt{xy}\,\frac{dy}{dx}=1$$

is separable. The usual textbook treatment assumes $x>0$ and $y>0$, so that $\sqrt{xy}=\sqrt{x}\sqrt{y}$. Under that assumption, separate the variables and integrate to obtain the family of solutions below. If the equation is intended over the reals without assuming positivity, there is also a branch with both $x$ and $y$ negative.

Solve on the positive branch

When $x>0$ and $y>0$, divide by $\sqrt{x}$ and multiply by $dx$ to separate the variables:

$$2\sqrt{y}\,dy=\frac{dx}{\sqrt{x}}.$$

Integrate both sides:

$$\int 2y^{1/2}\,dy=\int x^{-1/2}\,dx,$$

so

$$\frac{4}{3}y^{3/2}=2\sqrt{x}+C.$$

Absorbing the factor $3/4$ into the arbitrary constant gives the equivalent explicit form

$$y=\left(\frac{3}{2}\sqrt{x}+C_1\right)^{2/3}.$$

This formula is valid on an interval where $x>0$ and the quantity inside parentheses is positive. The positivity condition matters: this branch describes $y>0$, and the original equation is not defined as an equality at $y=0$ because its left side would be zero rather than one.

Use an initial condition

Suppose a solution is required to pass through $(1,1)$. Substituting this point into the integrated relation gives

$$\frac{4}{3}=2+C,$$

so $C=-2/3$. Therefore

$$\frac{4}{3}y^{3/2}=2\sqrt{x}-\frac{2}{3},$$

and hence

$$y=\left(\frac{3}{2}\sqrt{x}-\frac{1}{2}\right)^{2/3}.$$

For this positive branch, the expression in parentheses must be positive, which requires $x>1/9$. The endpoint $x=1/9$ is excluded because it would give $y=0$, where the differential equation cannot hold. The interval containing the initial point $x=1$ is therefore $(1/9,\infty)$.

As a check, let $A=\frac{3}{2}\sqrt{x}-\frac{1}{2}$. Then $y=A^{2/3}$ and

$$\frac{dy}{dx}=\frac{1}{2\sqrt{x}\,A^{1/3}}.$$

Since $\sqrt{xy}=\sqrt{x}\,A^{1/3}$ on this branch, substitution gives $2\sqrt{xy}\,dy/dx=1$.

What if $x$ and $y$ are negative?

For real values, the square root requires $xy\geq 0$. The equation itself rules out $xy=0$, so any solution interval must have $xy>0$: $x$ and $y$ have the same sign. If both are negative, you cannot use $\sqrt{xy}=\sqrt{x}\sqrt{y}$ with real square roots. Instead, use absolute values when separating:

$$2\sqrt{|y|}\,dy=\frac{dx}{\sqrt{|x|}}.$$

A compact way to express both real branches is to define $P(y)=\operatorname{sgn}(y)|y|^{3/2}$. Its derivative is $P'(y)=\frac{3}{2}\sqrt{|y|}$ for $y\ne 0$. Integration then gives

$$\operatorname{sgn}(y)|y|^{3/2} =\frac{3}{2}\operatorname{sgn}(x)\sqrt{|x|}+C.$$

For the negative branch, the right-hand side must be negative, just as $y$ is negative. This condition, along with $x<0$, restricts the interval on which the formula represents a solution. The positive branch above is recovered by taking $x>0$ and $y>0$.

Common mistakes

• Ignoring the domain: The equation cannot hold at $x=0$ or $y=0$. A solution must stay in a region where $xy>0$.
• Using the positive-branch factorization for negative values: For real variables, $\sqrt{xy}=\sqrt{x}\sqrt{y}$ is directly valid when both are nonnegative, not when both are negative. Use absolute values for the negative branch.
• Losing the constant or a factor: The integral of $2\sqrt{y}$ is $\frac{4}{3}y^{3/2}$, while the integral of $1/\sqrt{x}$ is $2\sqrt{x}$. Constants may be rescaled, but the coefficients from integration must be correct.
• Keeping an invalid endpoint: If the explicit formula reaches $y=0$, that endpoint cannot be included in a solution of the original equation.