A differential equation of the form \( M(x, y) dx + N(x, y) dy = 0 \) is said to be exact if there exists a function \( u(x, y) \) such that:
\[ du = M dx + N dy \]
Example:
\( y e^x dx + e^x dy = 0 \) is an exact equation since there exists a function \( y e^x \) such that:\[ d(y e^x) = y e^x dx + e^x dy \]
Theorem: The necessary and sufficient condition for the differential equation \( M dx + N dy = 0 \) to be exact is
\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]
Working Rule for Solving an Exact Differential Equation:
- Write the differential equation in the form \( M(x, y) dx + N(x, y) dy = 0 \).
- Find \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).
- If \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), then the given differential equation is exact.
- Write the general solution as
\[ \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} M dx + \int (\text{Terms of } N \text{ not containing } x) dy = c \]
1. Solve \( (2xy + y - \tan y) dx + (x^2 - x \tan^2 y + \sec^2 y) dy = 0 \)
Solution: The given equation is of the form \( M dx + N dy = 0 \),
\( M = 2xy + y - \tan y \) and \( N = x^2 - x \tan^2 y + \sec^2 y \)
Where \[ \frac{\partial M}{\partial y} = 2x + 1 - \sec^2 y \quad \text{and} \quad \frac{\partial N}{\partial x} = 2x - \tan^2 y = 2x + 1 - \sec^2 y \]
Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact.
The general solution is
\[ \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} (2xy + y - \tan y) dx + \int \sec^2 y \, dy = c \]
Or,
\[ 2y \int x \, dx + (y - \tan y) \int dx + \tan y = c \]
\[ x^2 y + (y - \tan y) x + \tan y = c \]
2. Solve \( (\cos x \tan y + \cos(x + y)) dx + (\sin x \sec^2 y + \cos(x + y)) dy = 0 \)
Solution: The given equation is of the form \( M dx + N dy = 0 \),
Where \( M = \cos x \tan y + \cos(x + y) \) and \( N = \sin x \sec^2 y + \cos(x + y) \)
\[ \frac{\partial M}{\partial y} = \cos x \sec^2 y - \sin(x + y) \quad \text{and} \quad \frac{\partial N}{\partial x} = \cos x \sec^2 y - \sin(x + y) \]
Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact.
The general solution is
\[ \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} (\cos x \tan y + \cos(x + y)) dx + \int(0) \, dy = c \]
\[ \tan y \int \cos x \, dx + \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} \cos(x + y) \, dx = c \]
\[ \sin x \tan y + \sin(x + y) = c \]
3. Solve \( (5x^4 + 3x^2 y^2 - 2xy^3) dx + (2x^3y - 3x^2 y^2 - 5y^4) dy = 0 \)
Solution: The given equation is of the form \( M dx + N dy = 0 \),
Where \( M = 5x^4 + 3x^2 y^2 - 2xy^3 \) and \( N = 2x^3y - 3x^2 y^2 - 5y^4 \)
\[ \frac{\partial M}{\partial y} = 6x^2 y - 6xy^2 \quad \text{and} \quad \frac{\partial N}{\partial x} = 6x^2 y - 6xy^2 \]
Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the given equation is exact.
The general solution is
\[ \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} (5x^4 + 3x^2 y^2 - 2xy^3) dx - \int 5y^5 \, dy = c \]
That is,
\[ 5\int x^4 dx + 3y^2 \int x^2 dx - 2y^3 \int x dx - 5 \int y^4 dy = c \]
\[ x^5 + x^3 y^2 - x^2 y^3 - y^5 = c \]
4. Solve \( (1 + e^{x/y}) dx + (1 - x/y) e^{x/y} dy = 0 \)
Solution: The given equation is of the form \( M dx + N dy = 0 \),
Where \( M = 1 + e^{x/y} \) and \( N = (1 - x/y) e^{x/y} \)
\[ \frac{\partial M}{\partial y} = e^{x/y} \left( -\frac{x}{y^2} \right) \quad \text{and} \quad \frac{\partial N}{\partial x} = (1 - x/y) e^{x/y} \left( \frac{1}{y} \right) - (1/y) e^{x/y} = e^{x/y} \left( -\frac{x}{y^2} \right) \]
Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the given equation is exact.
The general solution is
\[ \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} (1 + e^{x/y}) dx + \int(0) \, dy = c \]
\[ \Rightarrow \int dx + \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} e^{x/y} \, dx = c \]
\[ x + y e^{x/y} = c \]
Model Problems
Solve the following differential equations:
- \( \left[ y \left( 1 + \frac{1}{x} \right) + \cos y \right] dx + (x + \log x - x \sin y) dy = 0 \)
- \( \frac{dy}{dx} + \frac{y \cos x + \sin y + y}{\sin x + x \cos y + x} = 0 \)
- \( (y^2 e^{-x y^2} + 4x^3) dx + (2xy e^{-x y^2} - 3y^2) dy = 0 \)
- \( (1 + 2xy \cos x^2 - 2xy) dx + (\sin x^2 - x^2) dy = 0 \)
- \( (2x^2 + 3y^2 - 7) x dx + (3x^2 + 2y^2 - 8) y dy = 0 \)
- \( (e^y + 1) \cos x dx + e^y \sin x dy = 0 \)
- \( y \sin 2x dx - (1 + y^2 + \cos^2 x) dy = 0 \)
- \( (x^2 + 2y e^{2x}) dy + (2xy + 2y^2 e^{2x}) dx = 0 \)
Answers
- \( y(x + \log x) + x \cos y = c \)
- \( y \sin x + x(\sin y + y) = c \)
- \( e^{x y^2} + x^4 - y^3 = c \)
- \( y \sin x^2 - x^2 y + x = c \)
- \( x^4 + 3x^2 y^2 - 7x^2 + y^4 - 8y^2 = c \)
- \( (e^y + 1) \sin x = c \)
- \( 3y \cos 2x + 2y^3 + 6y = c \)
- \( x^2 y + y^2 e^{2x} = c \)