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Chapter 1: Problem 6

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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: \[ \phi(x, y) = y^2 x - x^2 + c_1(y) \\ \psi(x, y) = xy^2 + c_2(x) \]
03

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: \[ c_1(y) = F(x, y), \quad c_2(x) = -x^2 \]
04

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: \[ F(x, y) = x(y^2 - x) + C \]
05

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.

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