Variable separable equations

A first-order differential equation is variable separable if the right side factors purely into a function of $x$ times a function of $y$, with no term that mixes the two variables together:

where: $f(x)$ depends only on $x$ and $g(y)$ depends only on $y$; if the right side cannot be split this way (for example $x+y$, where the two variables are added rather than multiplied), the equation is not directly separable. Once recognised, rearrange so every $y$-term (including $dy$) sits on one side and every $x$-term (including $dx$) sits on the other, then integrate each side on its own:

where: $C$ is a single arbitrary constant of integration — only one is needed, because the two separate constants produced by integrating each side individually can always be combined into one overall constant on either side of the equation. This relation between $x$ and $y$ is the general solution; substituting a given initial condition $y(x_0) = y_0$ pins down $C$ and yields the particular solution. Not every first-order equation is separable as written, but some become separable after a substitution. If the right side is a function of a linear combination of $x$ and $y$, such as $\dfrac{dy}{dx} = f(ax+by+c)$, setting $v = ax+by+c$ converts it into a separable equation in $v$ and $x$, since $\dfrac{dv}{dx} = a + b\dfrac{dy}{dx}$ lets $\dfrac{dy}{dx}$ be replaced entirely in terms of $\dfrac{dv}{dx}$. Among the three first-order techniques in this syllabus — separable, homogeneous, linear — separable is the natural first check on any equation: the homogeneous method itself reduces to a separable equation after the substitution $y = vx$, so spotting separability early can save a great deal of algebra.

Solve $\dfrac{dy}{dx} = \dfrac{x^2+1}{y}$, given the initial condition $y(0) = 2$.

Step 1. Check the form and separate. The right side is $\dfrac{1}{y}\cdot(x^2+1)$, a product of a function of $y$ alone and a function of $x$ alone, so the equation is separable. Cross-multiplying, $y\,dy = (x^2+1)\,dx$.

Step 2. Integrate both sides. $\displaystyle\int y\,dy = \int (x^2+1)\,dx$ gives $\dfrac{y^2}{2} = \dfrac{x^3}{3} + x + C$, the general solution.

Step 3. Apply the initial condition. Substituting $x=0,\ y=2$: $\dfrac{4}{2} = 0+0+C$, so $C=2$.

Step 4. Write the particular solution and check it. $\dfrac{y^2}{2} = \dfrac{x^3}{3}+x+2$, equivalently $y^2 = \dfrac{2x^3}{3}+2x+4$. Differentiating the implicit form gives $y\,\dfrac{dy}{dx} = x^2+1$, i.e. $\dfrac{dy}{dx} = \dfrac{x^2+1}{y}$ — matching the original equation, so the solution checks out. Final answer: $y^2 = \dfrac{2x^3}{3}+2x+4$.

Competitive stretch

For $\dfrac{dy}{dx} = e^{x-y}$ with $y(0)=0$: rewrite $e^{x-y} = e^x e^{-y}$, separate as $e^y\,dy = e^x\,dx$, integrate to $e^y = e^x+C$, apply $x=0,y=0$ to get $C=0$, giving $e^y = e^x$, so $y=x$. The only new skill is splitting an exponential of a sum or difference before separating.

• Trying to separate an equation that isn't in product form. $\dfrac{dy}{dx} = x+y$ cannot be separated directly — the $x$ and $y$ are added, not multiplied — so it needs the linear-equation method instead.
• Writing two constants instead of one. Integrating each side gives its own constant, but they must be combined into a single overall $C$; keeping two separately is sloppy and confuses the step where the initial condition is applied.
• Losing constant solutions when dividing by $g(y)$. If $g(y)=0$ at some value $y=k$, then $y=k$ is itself a solution that gets divided away in Step 1; ISC problems rarely probe this, but it is worth naming.
• Forgetting to convert back after a reducing substitution. After solving in terms of $v = ax+by+c$, the final answer must be rewritten in terms of $x$ and $y$, not left in $v$.

Ask the tutor: Not sure whether your differential equation is separable, which substitution to try, or how to apply an initial condition correctly? Ask the AI tutor to walk through the structure and check each algebraic step with you.