Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 23 Marks
Question
Solve the differential equation: $\left( x ^3+ x ^2+ x +1\right) \frac{d y}{d x}=2 x ^2+ x$
✓
Answer
The given differential equation is,
$\left( x ^3+ x ^2+ x +1\right) \frac{d y}{d x}=2 x ^2+ x$
$\Rightarrow \frac{d y}{d x}=\frac{2 x^2+x}{x^3+x^2+x+1}$
$\Rightarrow d y=\frac{2 x^2+x}{(x+1)\left(x^2+1\right)} d x$
Integrating both sides, we get
$\int d y=\int\left\{\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}\right\} d x$
$\Rightarrow y=\int\left\{\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}\right\} d x$
Let $\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2+1}$
$\Rightarrow 2 x^2+x=A x^2+A+B x^2+B x+C x+C$
$\Rightarrow 2 x^2+x=(A+B) x^2+(B+C) x+(A+C)$
Comparing the coefficients on both sides, we get
$A + B = 2..... (i)$
$B + C = 1 ...(ii)$
$A + C = 0.... (iii)$
Solving $(i), (ii)$ and $(iii),$ we get
$A =\frac{1}{2}, B=\frac{3}{2}, C =-\frac{1}{2}$
$\therefore y=\frac{1}{2} \int \frac{1}{(x+1)} d x+\int \frac{\frac{3}{2} x-\frac{1}{2}}{x^2+1} d x$
$=\frac{1}{2} \int \frac{1}{(x+1)} d x+\frac{1}{2} \int \frac{3 x}{x^2+1} d x-\frac{1}{2} \int \frac{1}{x^2+1} d x$
$=\frac{1}{2} \int \frac{1}{(x+1)} d x+\frac{3}{4} \int \frac{2 x}{x^2+1} d x-\frac{1}{2} \int \frac{1}{x^2+1} d x$
$=\frac{1}{2} \log |x+1|+\frac{3}{4} \log \left|x^2+1\right|-\frac{1}{2} \tan ^{-1} x+C$
Hence, $y =\frac{1}{2} \log | x +1|+\frac{3}{4} \log \left| x ^2+1\right|-\frac{1}{2} \tan ^{-1} x + C$ is the solution to the given differential equation.
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