Solving Differential Equations by Exact Differential Equations

Solving Differential Equations by Exact Differential Equations:

1. Method Overview:

• Applicable to first-order ordinary differential equations (ODEs) in the form $M\left(x,y\right)dx+N\left(x,y\right)dy=0$.
• Checks for the existence of an integrating factor to make the equation exact, i.e., $\frac{\mathrm{\partial }M}{\mathrm{\partial }y}=\frac{\mathrm{\partial }N}{\mathrm{\partial }x}$.
2. Steps to Solve by Exact Differential Equations:

a. Check for Exactness: Given a differential equation in the form $M\left(x,y\right)dx+N\left(x,y\right)dy=0$, verify if $\frac{\mathrm{\partial }M}{\mathrm{\partial }y}=\frac{\mathrm{\partial }N}{\mathrm{\partial }x}$.

b. Find Integrating Factor: If the equation is not exact, find a suitable integrating factor $\mu \left(x,y\right)$ such that after multiplication by this factor, the equation becomes exact. The integrating factor $\mu \left(x,y\right)$ is obtained by $\mu ={e}^{\int \frac{{M}_{y}-{N}_{x}}{N}\text{\hspace{0.17em}}dx}$.

c. Multiply and Solve: Multiply the given equation by the integrating factor $\mu \left(x,y\right)$ and rearrange to obtain an equation that is now exact. Then, solve it using integration techniques.

3. Example: Solving a Differential Equation by Exact Differential Equations:

Consider the equation $\left(2xy+1\right)dx+\left({x}^{2}+2y\right)dy=0$.

a. Check for Exactness:

• Calculate $\frac{\mathrm{\partial }M}{\mathrm{\partial }y}=2x$ and $\frac{\mathrm{\partial }N}{\mathrm{\partial }x}=2x$. As they are equal, the equation is exact.

b. Solve the Exact Equation:

• Integrate $M\left(x,y\right)dx$ to find the function $f\left(x,y\right)$: $f\left(x,y\right)=\int \left(2xy+1\right)\text{\hspace{0.17em}}dx={x}^{2}y+x+g\left(y\right)$
• Take the partial derivative of $f\left(x,y\right)$ with respect to $y$ and set it equal to $N\left(x,y\right)$ to find $g\left(y\right)$: $\frac{\mathrm{\partial }f}{\mathrm{\partial }y}={x}^{2}+{g}^{\mathrm{\prime }}\left(y\right)={x}^{2}+2y$.
• Solve for $g\left(y\right)$ to get $g\left(y\right)={y}^{2}+C$.

c. Final Result:

• The solution to the differential equation $\left(2xy+1\right)dx+\left({x}^{2}+2y\right)dy=0$ by the method of exact differential equations is ${x}^{2}y+x+{y}^{2}+C=0$, where $C$ is the constant of integration.

Example: Solving a Non-Exact Differential Equation by Integrating Factor

Consider the non-exact differential equation ${y}^{\mathrm{\prime }}-\frac{y}{x}={e}^{-x}$.

1. Identify the Equation:

• The given equation is non-exact since $\frac{\mathrm{\partial }M}{\mathrm{\partial }y}\mathrm{\ne }\frac{\mathrm{\partial }N}{\mathrm{\partial }x}$.
2. Integrating Factor Method:

• To make the equation exact, introduce an integrating factor $\mu \left(x\right)$ by multiplying the entire equation.
3. Finding the Integrating Factor:

• For the equation ${y}^{\mathrm{\prime }}-\frac{y}{x}={e}^{-x}$, the integrating factor is $\mu \left(x\right)={e}^{\int -\frac{1}{x}\text{\hspace{0.17em}}dx}={e}^{-\mathrm{ln}\mathrm{\mid }x\mathrm{\mid }}=\frac{1}{\mathrm{\mid }x\mathrm{\mid }}$.
4. Multiply by Integrating Factor:

• Multiply the given equation ${y}^{\mathrm{\prime }}-\frac{y}{x}={e}^{-x}$ by the integrating factor $\frac{1}{\mathrm{\mid }x\mathrm{\mid }}$ to make it exact.

$\frac{1}{\mathrm{\mid }x\mathrm{\mid }}\left({y}^{\mathrm{\prime }}-\frac{y}{x}\right)=\frac{1}{\mathrm{\mid }x\mathrm{\mid }}\cdot {e}^{-x}$  $\frac{1}{x}{y}^{\mathrm{\prime }}-\frac{1}{{x}^{2}}y=\frac{1}{x}{e}^{-x}$

5. Solving the Exact Equation:

• Now, solve the transformed exact equation.
• Let $\frac{1}{x}{y}^{\mathrm{\prime }}-\frac{1}{{x}^{2}}y=\frac{1}{x}{e}^{-x}$ be the exact equation.

a. Integrate with Respect to $x$:

• Integrate both sides of the equation with respect to $x$.
• $\int \frac{1}{x}{y}^{\mathrm{\prime }}\text{\hspace{0.17em}}dx-\int \frac{1}{{x}^{2}}y\text{\hspace{0.17em}}dx=\int \frac{1}{x}{e}^{-x}\text{\hspace{0.17em}}dx$
• $\mathrm{ln}\mathrm{\mid }x\mathrm{\mid }\cdot y+\frac{1}{x}y=-{e}^{-x}+C$, where $C$ is the constant of integration.
6. Final Result:

• The solution to the non-exact differential equation ${y}^{\mathrm{\prime }}-\frac{y}{x}={e}^{-x}$ using the integrating factor method is $y=-x{e}^{-x}+Cx$, where $C$ is the constant of integration.