Question

Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\]  = x (x + 1) and passing through (1, 0).

Answer 1

We have, 
\[x\left( x + 1 \right)\frac{dy}{dx} - y = x\left( x + 1 \right)\]
\[ \Rightarrow \frac{dy}{dx} - \frac{y}{x\left( x + 1 \right)} = 1\]
\[\text{ Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get }\]
\[P = - \frac{1}{x\left( x + 1 \right)}\]
\[Q = 1\]
Now, 
\[I . F . = e^{- \int\frac{1}{x\left( x + 1 \right)}dx} \]
\[ = e^{- \int\frac{1}{x} - \frac{1}{x + 1}dx} \]
\[ = e^{- \log\left| \frac{x}{x + 1} \right|} \]
\[ = \frac{x + 1}{x} \]
So, the solution is given by
\[y \times I . F . = \int Q \times I . F . dx + C\]
\[ \Rightarrow \left( \frac{x + 1}{x} \right)y = \int\frac{x + 1}{x} dx + C\]
\[ \Rightarrow \left( \frac{x + 1}{x} \right)y = \int dx + \int\frac{1}{x}dx + C\]
\[ \Rightarrow \left( \frac{x + 1}{x} \right)y = x + \log \left| x \right| + C\]
\[\text{ Since the curve passes throught the point }\left( 1, 0 \right), \text{ it satisfies the equation of the curve . }\]
\[ \Rightarrow \left( \frac{1 + 1}{1} \right)0 = 1 + \log \left| 1 \right| + C\]
\[ \Rightarrow C = - 1\]
Putting the value of C in the equation of the curve, we get
\[\left( \frac{x + 1}{x} \right)y = x + \log \left| x \right| - 1\]
\[ \Rightarrow y = \frac{x}{x + 1}\left( x + \log \left| x \right| - 1 \right)\]