05. Exact Differential Equation
Dated: 08-11-2024
\[
\frac{dy}{dx} = f(x, y)
\]
\[
f(x,y) = -\frac{M(x,y)}{N(x,y)}
\]
\[
M(x,y)dx + N(x,y)dy = 0
\]
This equation is exact differential equation if the following condition is satisfied
\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
\]
\[
\frac{\partial F}{\partial x} = M(x,y), \quad \frac{\partial F}{\partial y} = N(x,y)
\]
Method of Solution
- Check the equation is
exact differential equation. - Write the system \(\frac{\partial F}{\partial x} = M(x,y), \quad \frac{\partial F}{\partial y} = N(x,y)\).
Integrate1 either the \(1^{\text{st}}\) equation with respect to \(x\) or \(2^{\text{nd}}\) with respect to \(y\). If we choose \(1^{\text{st}}\) equation then
\[
F(x,y) = \int M(x,y)dx + \theta(y)
\]
The \(\theta(y)\) is an arbitrary function.2
- Use the \(2^{\text{nd}}\) equation in step 2 and the equation from step 3 to find \(\theta^\prime(y)\).
\[
\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}\left(\int M(x,y)dx+\theta^{\prime}(y)\right) = N(x,y)
\]
\[
\theta^{\prime}(y) = N(x,y) - \frac{\partial}{\partial y}\int M(x,y)dx
\]
Integrate1 \(\theta^\prime(y)\) and write down thefunction2 \(F(x, y)\).- All solutions are given by implicit
function2
\[F(x, y) = C\]
- If we are given the
initial value problem3 then plugin the condition to find \(C\). - \(x\) should disappear from \(\theta^\prime(y)\) otherwise something is wrong.
Example
\[\frac {dy}{dx} = - \frac{3x^2y + 2}{x^3 + y}\]
\[
(3x^{2}y+2)dx+(x^{3}+y)dy=0
\]
\[
M=3x^{2}y+2 \text{ and } N=x^{3}+y
\]
Step 1
\[
\frac{\partial M}{\partial y} = 3x^{2}, \quad \frac{\partial N}{\partial x} = 3x^{2}
\]
\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
\]
Step 2
\[
\frac{\partial f}{\partial x} = 3x^{2}y+2 = M
\]
\[
\frac{\partial f}{\partial y} = x^{3}+y = N
\]
Step 3
Integrating2 \(1^{\text{st}}\) equation with respect to \(x\), we get
\[
f(x,y)=x^{3}y+2x+h(y)
\]
Step 4
Finding the derivative3 of the above function2 with respect to \(y\).
\[
\frac{\partial f}{\partial y} = x^{3} + h^{\prime}(y) = x^{3} + y = N
\]
Following function2 is independent of \(x\)
\[h^\prime(y) = y\]
and integrating1 it, we get
\[
h(y)=\frac{y^{2}}{2}
\]
Thus
\[
f(x,y)=x^{3}y+2x+\frac{y^{2}}{2}
\]
Step 5
\[
f(x,y)=c
\]
\[
x^{3}y+2x+\frac{y^{2}}{2}=c
\]
References
Read more about notations and symbols.