If $x^\alpha \frac{d y}{d x}=y^\beta(\gamma \log x+\delta \log y+1)$ is a homogeneous differential equation,…
If $x^\alpha \frac{d y}{d x}=y^\beta(\gamma \log x+\delta \log y+1)$ is a homogeneous differential equation, then
- $\alpha=\beta$ and $\gamma=-\delta$
- $\alpha=\beta$ and $\gamma=\delta$
- $\alpha \neq \beta$ and $\gamma=\delta$
- $\alpha \neq \beta$ and $\gamma \neq \delta$
Solution
Given, $x^\alpha \frac{d y}{d x}=y^\beta(\gamma \log x+\delta \log y+1)$
$\Rightarrow \frac{d y}{d x}=\frac{y^\beta}{x^\alpha}\left(\log x^\gamma \cdot y^\delta e\right)$
for homogeneous differential equation $\alpha=\beta$ and $\gamma=-\delta$.
Asked in: AP EAMCET 2023 (15 May Shift 2)
Practice more Differential Equations questions on Aicharya