x dx y − 2x dy 0 (differential equation Problems and solutions)
[latex]
\[
\textbf{Q1: Solve the differential equation:} \quad x \, dx + y – 2x \, dy = 0
\]
\[
\textbf{Step 1: Rearranging the terms}
\]
We start by grouping the terms involving \( dx \) and \( dy \) on opposite sides:
\[
x \, dx = 2x \, dy – y
\]
Now, rearrange to separate the variables:
\[
\frac{dx}{dy} = 2 – \frac{y}{x}
\]
\[
\textbf{Step 2: Separate the variables}
\]
To separate the variables, we rewrite the equation as:
\[
\frac{dx}{dy} = \frac{2x – y}{x}
\]
Now express it as:
\[
\frac{dx}{dy} = 2 – \frac{y}{x}
\]
\[
\textbf{Step 3: Integration}
\]
We can now integrate both sides of the equation. To do that, we’ll need to integrate the expression with respect to \( y \). First, let’s rewrite the equation in a more convenient form:
\[
x \, dx = (2x – y) \, dy
\]
Integrating both sides:
\[
\int x \, dx = \int (2x – y) \, dy
\]
On integrating:
\[
\frac{x^2}{2} = \int (2x \, dy – y \, dy)
\]
The integration yields:
\[
\frac{x^2}{2} = x^2 – \frac{y^2}{2} + C
\]
\[
\textbf{Step 4: Final solution}
\]
Simplifying the final equation:
\[
\frac{x^2}{2} = x^2 – \frac{y^2}{2} + C
\]
Multiply through by 2 to eliminate the fractions:
\[
x^2 = 2x^2 – y^2 + 2C
\]
Rearranging the terms:
\[
x^2 – 2x^2 + y^2 = 2C
\]
\[
-x^2 + y^2 = 2C
\]
Finally, the solution is:
\[
y^2 – x^2 = C
\]
This is the general solution to the differential equation \( x \, dx + y – 2x \, dy = 0 \), where \( C \) is a constant of integration.