Kit Library / Mathematics / Differential Equations
⚡ Topic Learning KitShared by a Veda learner · Generated with Veda AI
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Check exactness first — it saves hours.
An equation $M\,dx + N\,dy = 0$ is exact when $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Only then can you find a potential function whose total differential equals the equation.
↳ If the cross partials match, the equation is exact and a single potential function $F(x,y)=C$ solves it.
Exact means a perfect differential exists.
↳ Exactness guarantees that the left side is the total differential of some $F(x,y)$.
Four steps from equation to solution.
↳ The method always reduces to two integrations and one differentiation — no guessing required.
This topic covers the exact differential equation method applied to (1 + e^(x/y))dx + e^(x/y)(1 - x/y)dy = 0. It walks through the exactness test using partial derivatives, integration of M with respect to x, and determi…
~15 min · full explanation, examples & memory tricks in the app
19 questions laddered from warm-up to topper-level, each with an explanation. A taste:
A first-order differential equation is written in the standard form $M\,dx + N\,dy = 0$. Which condition must hold for it to be exact?
Beginner$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$
The exactness test for $M\,dx + N\,dy = 0$ is $\partial M/\partial y = \partial N/\partial x$. The other options mix up which variable each function is differentiated with respect to.
For the equation $(1 + e^{x/y})dx + e^{x/y}(1 - x/y)dy = 0$, what is $M$?
Beginner$1 + e^{x/y}$
In the form $M\,dx + N\,dy = 0$, $M$ is the coefficient of $dx$, which is $1 + e^{x/y}$.
For the same equation, what is $N$?
Beginner$e^{x/y}(1 - x/y)$
$N$ is the coefficient of $dy$, which is $e^{x/y}(1 - x/y)$.
Compute $\dfrac{\partial M}{\partial y}$ for $M = 1 + e^{x/y}$.
Beginner$-\dfrac{x}{y^2}e^{x/y}$
Differentiating $e^{x/y}$ with respect to $y$ gives $e^{x/y} \cdot (-x/y^2) = -\dfrac{x}{y^2}e^{x/y}$. The constant $1$ differentiates to $0$.
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