Differential equations with variables separable

A first-order differential equation is said to have variables separable when it can be written as a product of a pure function of $x$ and a pure function of $y$:

where: $h(x)$ depends only on $x$, $g(y)$ depends only on $y$, and $\dfrac{dy}{dx}$ is the rate of change linking them. Recognising this shape is the whole game — if the right-hand side cannot be factored this way, the method simply does not apply. Once the shape is confirmed, divide both sides by $g(y)$ (assuming $g(y) \neq 0$) and multiply by $dx$ to physically separate the differentials:

where: each side now contains only one variable together with its own differential, so the two sides can be treated as completely independent integration problems. Integrating both sides independently gives the general solution:

where: $C$ is a single arbitrary constant. Each side technically produces its own constant of integration, but since only their difference matters, the two collapse into one net constant $C$ written on either side — never both. The figure below shows the separation as a literal rearrangement: terms migrate across the equals sign until $y$-stuff and $x$-stuff sit on opposite banks, ready to be integrated.

Not every equation qualifies. When $x$ and $y$ are genuinely mixed — a sum like $x + y$, for instance — the right-hand side cannot be split into a pure-$x$ factor times a pure-$y$ factor, and a different technique is needed. That is precisely the motivation for the homogeneous-equation method taught next: a substitution ($y = vx$) that turns certain non-separable equations into separable ones, bringing them back within reach of this same integration approach.

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$, given that $y = 4$ when $x = 3$.

Step 1. Confirm the equation is separable: $\dfrac{dy}{dx} = x \cdot \dfrac{1}{y}$ is a pure function of $x$ times a pure function of $y$, so the method applies directly.

Step 2. Separate the variables by cross-multiplying:

Step 3. Integrate both sides:

Step 4. Simplify to the standard general-solution form. Multiplying through by $2$ and writing $C = 2C_1$:

Step 5. Apply the initial condition $x = 3,\ y = 4$ to pin down $C$:

Step 6. Write the particular solution:

Check: differentiating $y^2 - x^2 = 7$ implicitly gives $2y\dfrac{dy}{dx} - 2x = 0 \implies \dfrac{dy}{dx} = \dfrac{x}{y}$, which matches the original equation exactly, and $x=3,y=4$ satisfies $16-9=7$. The particular solution is $y^2 - x^2 = 7$.

Competitive stretch

Solve $\dfrac{dy}{dx} = e^{x-y}$.

Step 1. Rewrite using laws of exponents: $\dfrac{dy}{dx} = e^x \cdot e^{-y}$, which is separable since it is a pure-$x$ factor times a pure-$y$ factor.

Step 2. Separate by multiplying both sides by $e^{y}$: $e^{y}\,dy = e^{x}\,dx$.

Step 3. Integrate: $\displaystyle\int e^{y}\,dy = \int e^{x}\,dx \implies e^{y} = e^{x} + C$.

The general solution is $e^{y} = e^{x} + C$, equivalently $y = \ln(e^x + C)$ wherever $e^x + C > 0$.

• Dividing by $g(y)$ without noting where it could be zero. The separation step assumes $g(y) \neq 0$; CBSE problems rarely test this edge case explicitly, but a careful answer flags it.
• Writing two constants instead of combining them into one. Integrating each side independently technically produces $C_1$ and $C_2$ — these must be merged into a single $C$ in the final general solution, not left floating on both sides.
• Forcing a separation that is not algebraically valid. An equation like $\dfrac{dy}{dx} = x + y$ cannot be split into a pure function of $x$ times a pure function of $y$; it needs the homogeneous-equation substitution instead, not a forced separation.
• Sign or algebra slips while rearranging exponential or fractional terms. A misplaced sign while moving a term across the equals sign silently changes the entire family of solutions.

Ask the tutor: if you're unsure whether a given equation is truly separable, or how to separate a trickier expression mixing exponentials or trig terms, ask your AI tutor to walk through the separation step before you integrate.