The general solution of differential equation $(y^2 - x^2)dx = xy dy$ ($x \neq 0$) is ______.
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To easily verify if a differential equation is homogeneous, replace $x$ with $kx$ and $y$ with $ky$. If all $k$'s cancel out perfectly, the equation is homogeneous and the substitution $y=vx$ will always work.
$2x^2 \log x + y^2 + 2cx^2 = 0$, where $c$ is the constant of integration
$2x^2 \log x - y^2 + 2cx^2 = 0$, where $c$ is the constant of integration
$x^2 \log x + y^2 + 2cx^2 = 0$, where $c$ is the constant of integration
$x^2 \log x - y^2 + 2cx^2 = 0$, where $c$ is the constant of integration
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The Correct Option isA
Solution and Explanation
Step 1: Understanding the Concept:
This is a homogeneous differential equation because the degrees of all terms in $x$ and $y$ are the same (degree 2). We substitute $y = vx$. Step 2: Formula Application:
$\frac{dy}{dx} = v + x \frac{dv}{dx}$.
The equation is $\frac{dy}{dx} = \frac{y^2 - x^2}{xy}$. Step 3: Explanation:
$v + x \frac{dv}{dx} = \frac{v^2x^2 - x^2}{x(vx)} = \frac{v^2 - 1}{v}$
$x \frac{dv}{dx} = \frac{v^2 - 1}{v} - v = \frac{v^2 - 1 - v^2}{v} = -\frac{1}{v}$
Integrating: $\int v \, dv = -\int \frac{1}{x} \, dx$
$\frac{v^2}{2} = -\log x - c \implies \frac{y^2}{2x^2} + \log x + c = 0$
Multiply by $2x^2$: $y^2 + 2x^2 \log x + 2cx^2 = 0$. Step 4: Final Answer:
The solution is $2x^2 \log x + y^2 + 2cx^2 = 0$.
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