Problem 6 - Differential Equations and Linear Algebra

Question

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

Step-by-Step Solution

Answer

The general solution to the given exact differential equation is: \[F(x, y) = x(y^2 - x) + C\] where C is an arbitrary constant.

Step 1: Check for exactness

First, we need to determine if the given differential equation is exact. An exact differential equation can be expressed in the form \(M(x, y) dx + N(x, y) dy = 0\). In this case, we can identify the functions: \[ M(x, y) = y^2 - 2x, \quad N(x, y) = 2xy \] For a differential equation to be exact, the condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) must be satisfied. Let's check the partial derivatives: \[ \frac{\partial M}{\partial y} = 2y, \quad \frac{\partial N}{\partial x} = 2y \] Since both the partial derivatives are equal, the given differential equation is exact.

Step 2: Integrate M and N

Now, we will integrate both functions M and N with respect to x and y, respectively: \[ \int M(x, y) dx = \int (y^2 - 2x) dx \\ \int N(x, y) dy = \int (2xy) dy \] Integrating both expressions, we obtain: \[ \phi(x, y) = y^2 x - x^2 + c_1(y) \\ \psi(x, y) = xy^2 + c_2(x) \]

Step 3: Determine the arbitrary functions

We will now determine the missing functions \(c_1(y)\) and \(c_2(x)\) by combining both expressions for \(\phi(x, y)\) and \(\psi(x, y)\): \[ \phi(x, y) = \psi(x, y) + F(x, y) \] Substituting the expressions for \(\phi(x, y)\) and \(\psi(x, y)\), we have: \[ y^2 x - x^2 + c_1(y) = xy^2 + c_2(x) + F(x, y) \] Comparing the terms, we determine the missing functions: \[ c_1(y) = F(x, y), \quad c_2(x) = -x^2 \]

Step 4: Write the combined solution function

Now that we have the arbitrary functions, we can write the general solution for the differential equation: \[ \phi(x, y) = y^2 x - x^2 + F(x, y) \] Simplifying, \[ F(x, y) = x(y^2 - x) + c(y) \\ \] Since the general solution only has one arbitrary constant, c(y), we can rewrite it as a constant C: \[ F(x, y) = x(y^2 - x) + C \]

Step 5: Final solution

The final solution to the given exact differential equation is: \[F(x, y) = x(y^2 - x) + C\] This is the general solution to the given differential equation, where C is an arbitrary constant.