In Calculus 3, we are often interested in level curves of the form \(F(t,y)=C\text{,}\) where \(C\) is an arbitrary constant. If we differentiate with respect to \(t\text{,}\) we get \(\dfrac{\mathrm{d}}{\mathrm{d}t} F(t,y) = \dfrac{\mathrm{d}}{\mathrm{d}t} C\) which simplifies (with "partial derivative" notation) to \(\dfrac{\partial F}{\partial t} + \dfrac{\partial F}{\partial y} \dfrac{\mathrm{d}y}{\mathrm{d}t}=0\text{.}\) Solving this for \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) then yields the differential equation
We use (1.12.2) to look for (implicit) solutions of the form \(F(t,y)=C\text{.}\)
Definition1.12.1.Exact equations.
The differential form (1.12.2) is said to be exact in a region \(R\) in the plane if there exists a function \(F\) such that \(\dfrac{\partial F}{\partial t}=M\) and \(\dfrac{\partial F}{\partial y}=N\) for all \((t,y) \in R\text{.}\)
Suppose \(M\) and \(N\) have continuous first-order partial derivatives in a region \(R\) in the plane. Then (1.12.2) is exact on \(R\) if and only if \(\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial t}\) for all \((t,y) \in R\text{.}\)
then the function \(\mu\) is called an integrating factor for (1.12.2). When using such an integrating factor, it is possible to gain or lose solutions.
If \(\dfrac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial t}}{N}\) is continuous and depends only on \(t\text{,}\) then we can take the integrating factor to be
If \(\dfrac{\frac{\partial N}{\partial t} - \frac{\partial M}{\partial y}}{M}\) is continuous and depends only on \(y\text{,}\) then we can take the integrating factor to be
Show that this equation becomes exact when both sides are multiplied by the integrating factor \(\mu(t,y)=\dfrac{t^2}{y^2}\text{.}\) Find an implicit solution.