Question

\[\left( x^3 + x^2 + x + 1 \right)\frac{dy}{dx} = 2 x^2 + x\]

Answer 1

We have, 
\[\left( x^3 + x^2 + x + 1 \right)\frac{dy}{dx} = 2 x^2 + x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2 x^2 + x}{x^3 + x^2 + x + 1}\]
\[ \Rightarrow dy = \frac{2 x^2 + x}{\left( x + 1 \right)\left( x^2 + 1 \right)}dx\]
Integrating both sides, we get
\[\int dy = \int\left\{ \frac{2 x^2 + x}{\left( x + 1 \right)\left( x^2 + 1 \right)} \right\}dx\]
\[ \Rightarrow y = \int\left\{ \frac{2 x^2 + x}{\left( x + 1 \right)\left( x^2 + 1 \right)} \right\}dx\]
\[\text{ Let }\frac{2 x^2 + x}{\left( x + 1 \right)\left( x^2 + 1 \right)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1}\]
\[ \Rightarrow 2 x^2 + x = A x^2 + A + B x^2 + Bx + Cx + C\]
\[ \Rightarrow 2 x^2 + x = \left( A + B \right) x^2 + \left( B + C \right)x + \left( A + C \right)\]
Comparing the coefficients on both sides, we get
\[A + B = 2 . . . . . \left( 1 \right)\]
\[B + C = 1 . . . . . \left( 2 \right)\]
\[A + C = 0 . . . . . \left( 3 \right)\]
\[\text{ Solving }\left( 1 \right), \left( 2 \right)\text{ and }\left( 3 \right),\text{ we get }\]
\[A = \frac{1}{2}\]
\[B = \frac{3}{2}\]
\[C = - \frac{1}{2}\]
\[ \therefore y = \frac{1}{2}\int\frac{1}{\left( x + 1 \right)}dx + \int\frac{\frac{3}{2}x - \frac{1}{2}}{x^2 + 1} dx\]
\[ = \frac{1}{2}\int\frac{1}{\left( x + 1 \right)}dx + \frac{1}{2}\int\frac{3x}{x^2 + 1}dx - \frac{1}{2}\int\frac{1}{x^2 + 1}dx\]
\[ = \frac{1}{2}\int\frac{1}{\left( x + 1 \right)}dx + \frac{3}{4}\int\frac{2x}{x^2 + 1}dx - \frac{1}{2}\int\frac{1}{x^2 + 1}dx\]
\[ = \frac{1}{2}\log\left| x + 1 \right| + \frac{3}{4}\log\left| x^2 + 1 \right| - \frac{1}{2} \tan^{- 1} x + C\]
\[\text{ Hence, }y = \frac{1}{2}\log\left| x + 1 \right| + \frac{3}{4}\log\left| x^2 + 1 \right| - \frac{1}{2} \tan^{- 1} x +\text{ C is the solution to the given differential equation }.\]