Question

Solve the following differential equation:-

(1 + x²) dy + 2xy dx = cot x dx

Answer 1

We have,

\[\left( 1 + x^2 \right)dy + 2xy dx = \cot x dx\]

\[ \Rightarrow \frac{dy}{dx} + \frac{2x}{\left( 1 + x^2 \right)}y = \frac{\cot x}{\left( 1 + x^2 \right)}\]

\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]

\[P = \frac{2x}{\left( 1 + x^2 \right)}\]

\[Q = \frac{\cot x}{\left( 1 + x^2 \right)}\]

Now,

\[I.F. = e^{\int\frac{2x}{\left( 1 + x^2 \right)}dx} \]

\[ = e^{\log\left| 1 + x^2 \right|}\]

\[ = 1 + x^2 \]

So, the solution is given by

\[y \times I . F . = \int Q \times I . F . dx + C\]

\[ \Rightarrow y\left( 1 + x^2 \right) = \int\left[ \frac{\cot x}{\left( 1 + x^2 \right)} \times \left( 1 + x^2 \right) \right] dx + C\]

\[ \Rightarrow y\left( 1 + x^2 \right) = \int\cot x dx + C\]

\[ \Rightarrow y\left( 1 + x^2 \right) = \log \left| \sin x \right| + C\]

\[ \Rightarrow y = \left( 1 + x^2 \right)^{- 1} \log \sin x + C \left( 1 + x^2 \right)^{- 1}\]