First-Order Equations: Quick Revision

A brief note on First-Order Equations. Read it before a test. Try each question at the end before you open its answer.

Separable equations

• Form: $\dfrac{dy}{dx} = f(x)\,g(y)$. NCERT says "variable separable". A-Level, IB and AP say "separable".
• Method: $\dfrac{dy}{g(y)} = f(x)\,dx$. Integrate both sides, with one constant, then use the point.
• Lost solutions: every $y$ with $g(y) = 0$ gives a constant solution. Check whether the general solution has it.

Homogeneous equations

• Form: $\dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right)$. The top and the bottom are homogeneous of the same degree.
• Method: put $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Then $x\dfrac{dv}{dx} = F(v) - v$ separates.
• Shortcut: use $x = vy$ when the equation is simpler for $\dfrac{dx}{dy}$.
• The word: A-Level Further Maths uses "homogeneous" for a different idea (zero on the right side).

Linear equations

• Form: $\dfrac{dy}{dx} + Py = Q$. The coefficient of $\dfrac{dy}{dx}$ must be $1$.
• Integrating factor: $I = e^{\int P\,dx}$, and $I\,y = \int IQ\,dx + C$.
• Other form: $\dfrac{dx}{dy} + P_1x = Q_1$ uses $e^{\int P_1\,dy}$.
• Shape of the answer: transient plus steady state. The transient fades.

A worked example

Solve $\dfrac{dy}{dx} = \dfrac{y}{x} + \dfrac{x}{y}$ for $x > 0$ and $y > 0$.

1. Check. The right side is $v + \dfrac{1}{v}$ with $v = \dfrac{y}{x}$. The equation is homogeneous.
2. Substitute. $v + x\dfrac{dv}{dx} = v + \dfrac{1}{v}$, so $v\,dv = \dfrac{dx}{x}$.
3. Integrate. $\dfrac{v^2}{2} = \ln x + c$. Put $v = \dfrac{y}{x}$: $y^2 = 2x^2(\ln x + C)$.

Practice with new numbers

A program to try

The program checks that $y = 6 - \dfrac{2}{x}$ solves $x\dfrac{dy}{dx} + y = 6$ at four points.

Mistakes to avoid

• Dividing by $g(y)$ and forgetting its zeros. Constant solutions can be lost.
• Forgetting $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Treating $v$ as a constant drops a term.
• Using the factor before the equation is in standard form. Divide by the coefficient of $\dfrac{dy}{dx}$ first.
• Forgetting to divide $C$ by the integrating factor.

Check yourself

1. Solve $\dfrac{dy}{dx} = 3x^2y^2$.

2. Is $\dfrac{dy}{dx} = \dfrac{x + 3y}{2x - y}$ homogeneous?

3. Find the integrating factor of $\dfrac{dy}{dx} + 2y = e^{3x}$.

4. Find the constant solutions of $\dfrac{dy}{dx} = y(3 - y)$.

5. Solve $\dfrac{dy}{dx} + y = 4$ with $y = 0$ at $x = 0$.

Sources

• OpenStax, Calculus Volume 2, Chapter 4, sections 4.3 and 4.5.
• NCERT, Mathematics Part II for Class XII, Chapter 9, the sections on methods of solving first-order, first-degree equations.