Variable-Separable Differential Equations

An equation is variable-separable if it can be written in the form $\dfrac{dy}{dx} = f(x)\,g(y)$, i.e. the right side factors into an x-only part times a y-only part. Then:

• Separate: $\dfrac{dy}{g(y)} = f(x)\,dx$ — divide across so each side holds one variable only.
• Integrate both sides once: $\displaystyle\int \dfrac{dy}{g(y)} = \int f(x)\,dx + c$.
• One arbitrary constant for the whole (first-order) equation — never one per side.

The number of arbitrary constants in the general solution equals the ORDER of the equation, so a first-order equation carries exactly one.

Once separated, reach for the elementary integrals:

• $\displaystyle\int \dfrac{dy}{y} = \log y$, $\displaystyle\int y\,dy = \dfrac{y^2}{2}$, $\displaystyle\int \dfrac{dx}{x^2} = -\dfrac{1}{x}$.
• Absorbing constants into $\log c$ turns $\log y = 2\log x + \log c$ into the clean family $y = cx^2$.
• A first-order linear-looking equation like $x\dfrac{dy}{dx} = 2y$ is really separable: $\dfrac{dy}{y} = 2\dfrac{dx}{x}$.

Procedure for an initial-value problem (IVP):

• Separate and integrate to the general solution with its arbitrary constant $c$.
• Substitute the given $(x_0, y_0)$ to solve for $c$.
• Substitute $c$ back, then evaluate at the requested point.

A very common MHT-CET shape is $(2+\sin x)\dfrac{dy}{dx} + (y+1)\cos x = 0$: separating gives $\dfrac{dy}{y+1} = -\dfrac{\cos x}{2+\sin x}\,dx$, so $\log(y+1) = -\log(2+\sin x) + c$, i.e. $(y+1)(2+\sin x) = k$.

Two recurring disguises:

• Log of the derivative: $\log\!\big(\tfrac{dy}{dx}\big) = ax + by \Rightarrow \tfrac{dy}{dx} = e^{ax}e^{by} \Rightarrow e^{-by}\,dy = e^{ax}\,dx$, giving $a\,e^{-by} + b\,e^{ax} = c_1$.
• Exponential factor on the RHS: $\tfrac{dy}{dx} = 2xy\,e^{x^2} \Rightarrow \dfrac{dy}{y} = 2x\,e^{x^2}\,dx$; put $u = x^2$ so the x-side is $\int e^u\,du = e^{x^2}$, giving $\log y = e^{x^2} + \log c$, i.e. $y = c\,e^{e^{x^2}}$.
• The product form $e^{\,y-x}\tfrac{dy}{dx} = \dfrac{y(\sin x+\cos x)}{1+y\log y}$ rearranges to $\dfrac{e^y(1+y\log y)}{y}\,dy = e^x(\sin x+\cos x)\,dx$, and the standard trick $\int e^x\big(f+f'\big)\,dx = e^x f$ collapses the RHS to $e^x\sin x$, giving $e^y\log y = e^x\sin x + c$.

Trig separables split into standard log-integrals:

• $\displaystyle\int \dfrac{\cos x}{1+\sin x}\,dx = \log(1+\sin x)$, $\displaystyle\int \dfrac{\sin y}{1+\cos y}\,dy = -\log(1+\cos y)$ — both are $\int \tfrac{f'}{f}$.
• $\dfrac{dy}{dx} = \cot x\cot y \Rightarrow \tan y\,dy = \cot x\,dx \Rightarrow -\log\cos y = \log\sin x - \log c$, i.e. $\sin x = c\sec y$.
• Product-to-sum first: $\sin\tfrac{x-y}{2} - \sin\tfrac{x+y}{2} = -2\cos\tfrac{x}{2}\sin\tfrac{y}{2}$, which then separates as $\csc\tfrac{y}{2}\,dy = -2\cos\tfrac{x}{2}\,dx$.

The standard integrals that appear here:

• $\displaystyle\int \dfrac{dy}{1+y^2} = \tan^{-1}y$; combining $\tan^{-1}y - \tan^{-1}x = \tan^{-1}c$ gives $\dfrac{y-x}{1+xy} = c$.
• $\displaystyle\int \dfrac{y\,dy}{\sqrt{k^2-y^2}} = -\sqrt{k^2-y^2}$, so $\dfrac{y\,dy}{\sqrt{k^2-y^2}} = \pm dx$ integrates to $x^2 + y^2 = k^2$ — a family of circles.
• $y\dfrac{dy}{dx} = a - x$ integrates to $x^2 + y^2 - 2ax - 2c = 0$: circles with centre $(a,0)$, radius $\sqrt{a^2+2c}$.

When $\dfrac{dy}{dx} = f(x)$, the solution is simply $y = \int f(x)\,dx + c$. Useful setups:

• Simplify first: $\dfrac{dy}{dx} = \dfrac{3e^{2x}+3e^{4x}}{e^x+e^{-x}} = \dfrac{3e^{2x}(1+e^{2x})}{e^{-x}(e^{2x}+1)} = 3e^{3x}$, so $y = e^{3x} + c$.
• Polynomial division: $(x+2)\dfrac{dy}{dx} = x^2+4x-9$ gives $\dfrac{dy}{dx} = (x+2) - \dfrac{13}{x+2}$, which integrates to $y = \tfrac{(x+2)^2}{2} - 13\log|x+2| + c$.
• A constant derivative from an implicit relation: $\cos\!\big(\tfrac{dy}{dx}\big) = 7 \Rightarrow \tfrac{dy}{dx} = \cos^{-1}7$ (a constant), so $y = (\cos^{-1}7)x + c$.