Question

The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by

विकल्प

• y = xe^x + C

• x = ye^x

• y = x + C

• xy = e^x + C

Answer 1

y = xe^{x + C}

 

We have,
\[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y\left( x + 1 \right)}{x}\]
\[ \Rightarrow \frac{dy}{y} = \frac{\left( x + 1 \right)}{x}dx\]
Integrating both sides, we get
\[\int\frac{dy}{y} = \int\frac{\left( x + 1 \right)}{x}dx\]
\[ \Rightarrow \int\frac{dy}{y} = \int dx + \int\frac{1}{x}dx\]
\[ \Rightarrow \log y = x + \log x + C\]
\[ \Rightarrow \log y - \log x = x + C\]
\[ \Rightarrow \log \left( \frac{y}{x} \right) = x + C\]
\[ \Rightarrow \frac{y}{x} = e^{x + C} \]
\[ \Rightarrow y = x e^{x + C}\]