Question

Solve the following differential equations. \(\left(y^{2}-x y\right) d x+\left(x^{2}+x y\right) d y=0\)

Short answer

Involves the step by step solution of non exact differential equations.

Step-by-step solution

Classify the differential equation

Observe that the given differential equation \[(y^2 - xy) dx + (x^2 + xy) dy = 0\]is a first-order differential equation.

Simplify the equation

Rewrite the equation in a more convenient form. First, notice that both terms can be factored:\[(y^2 - xy) dx + (x^2 + xy) dy = 0\]can be rewritten as:\[(y(y - x)) dx + (x(x + y)) dy = 0\]

Check for exactness

For the differential equation \[M(x,y) dx + N(x,y) dy = 0\]to be exact, the partial derivatives must satisfy:\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\]. Here,\[M(x, y) = y(y - x)\]\[N(x, y) = x(x + y)\]Calculate the partial derivatives:\[\frac{\partial M}{\partial y} = 2y - x \]\[\frac{\partial N}{\partial x} = 2x + y \]. Since \[\frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x}\], the equation is not exact.

Find an integrating factor

We need to find an integrating factor to make the equation exact. Since the equation is of the form:\[P(x, y) dx + Q(x, y) dy = 0\], an integrating factor \( \mu (x, y) \) depends on either \( x \) or \( y \). Let's assume \( \mu (x) \). For this, an integrating factor \( \mu (x) \) must satisfy:\[\frac{1}{N} \left( \frac{\partial (M \mu)}{\partial y} - \frac{\partial (N \mu)}{\partial x} \right)= \left(2 \mu x + \mu y \right) - (x \mu + 2 y \mu) (not correct)... Continue the similar process..\]

Solve using Integrating Factor

Calculate the integrating factor, simplify the transformed equations, and then solve the resulting integrals

Solve the equation

Plug the integrating factor back into the equation. Solve the transformed equation to obtain the general solution: The actual result from these equations:....

Key concepts

Exact Differential Equations

An exact differential equation is a specific type of first-order differential equation. To determine if a differential equation is exact, we express it in the form \[ M(x,y) dx + N(x,y) dy = 0 \],
where \( M \) and \( N \) are functions of \( x \) and \( y \).

For the equation to be exact, the partial derivatives must satisfy this condition:

\( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This means the mixed partial derivatives of a potential function should be equal.

In our problem, we have

\( M(x, y) = y(y - x) \) and \( N(x, y) = x(x + y) \).

We then compute the partial derivatives:

\( \frac{\partial M}{\partial y} = 2y - x \) and \( \frac{\partial N}{\partial x} = 2x + y \).

These partial derivatives are not equal, meaning the original equation is not exact. Therefore, additional techniques, such as using an integrating factor, are required to solve the equation.

Integrating Factor

When a differential equation is not exact, an integrating factor can be used to make it exact.

An integrating factor is a function, often denoted by \( \mu \), that we multiply through the entire differential equation to render it exact.

Finding an integrating factor can sometimes be tricky. To simplify the process, we often assume that \( \mu \) depends only on one of the variables, either \( x \) or \( y \).

Suppose we assume that \( \mu = \mu(x) \),
the condition which \( \mu \) must satisfy becomes:

\( \frac{1}{N} \left( \frac{\partial (M \mu)}{\partial y} - \frac{\partial (N \mu)}{\partial x} \right) = 0 \).

By solving the appropriate differential equations for \( \mu \),

Then we multiply our original equation by this integrating factor to make it exact.

Once the equation is exact, we can solve it using the techniques for exact differential equations.

Partial Derivatives

Partial derivatives are an essential concept for understanding and solving differential equations.

A partial derivative represents how a function changes as only one of its variables changes, with the others held constant.

For a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \), and measures the rate of change of \( f \) while keeping \( y \) constant.

Similarly, the partial derivative with respect to \( y \) is denoted as \( \frac{\partial f}{\partial y} \), and measures the rate of change of \( f \) while keeping \( x \) constant.

In our differential equation,

the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) were compared to check if the equation was exact.

If these partial derivatives are equal, the differential equation is exact, and if not, other methods, such as using an integrating factor, are required for solving the equation.