Elementary Differential Equations

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Exact Equations

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Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{-25y^2 + 4x + 10y + 14}{50xy - 10x - 52y + 12} \\ y(-1) &= -4 \end{align}$$ This can be written as an exact equation. First we find the general solution following the paradigm.

We write the equation in the standard form, M dx + N dy = 0. $$ (25y^2 - 4x - 10y - 14) dx + (50xy - 10x - 52y + 12) dy = 0 $$
We test for exactness. $$\frac{\partial}{\partial y}\left(25y^2 - 4x - 10y - 14\right) = 50y - 10 = \frac{\partial}{\partial x}\left(50xy - 10x - 52y + 12\right) $$ so the equation is exact.
Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= 25y^2 - 4x - 10y - 14\\ \frac{\partial F}{\partial y} &= 50xy - 10x - 52y + 12 \end{align}$$
Integrate the first partial differential equation. $$ F(x,y) = \int (25y^2 - 4x - 10y - 14)\,\partial x = 25xy^2 - 2x^2 - 10xy - 14x + C(y) $$
Integrate the second partial differential equation. $$ F(x,y) = \int (50xy - 10x - 52y + 12)\,\partial y = 25xy^2 - 10xy - 26y^2 + 12y + \tilde{C}(x) $$
Equate the expressions for F(x,y).
Matching the expressions up, we find $C(y) = -26y^2 + 12y$ and $ \tilde{C}(x) = -2x^2 - 14x. $ So $$ F(x,y) = 25xy^2 - 26y^2 - 10xy - 2x^2 - 14x + 12y. $$
The solution is $F(x,y) = K.$ $$ 25xy^2 - 26y^2 - 10xy - 2x^2 - 14x + 12y = K $$

Now we plug in the initial values $x = -1$ and $y = -4$ and solve for $K = -892$. So the solution to the initial value problem is $$ 25xy^2 - 26y^2 - 10xy - 2x^2 - 14x + 12y = -892 $$ You may reload this page to generate additional examples.