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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
For each of the differential equations in find a particular solution satisfying the given condition:
$(\text{x}^3+\text{x}^2+\text{x}+1) \frac{\text{dy}}{\text{dx}} = 2\text{x}^2+\text{x; y} =1 \ \text{when x = 0}$
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Answer

The given differential equation is
$(\text{x}^3+\text{x}^2+\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{x}^2+\text{x}$
$\text{or} \ \ \ \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^2+\text{x}}{\text{x}^3+\text{x}^2+\text{x}+1}$
$\text{or} \ \ \ \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^2+\text{x}}{\text{x}^2(\text{x}+1)+1(\text{x}+1)}$
$\text{or} \ \ \ \ \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}$
Separting the variables, we get,
$\text{dy}=\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\text{dx}$
Integrating, $\int\text{dy=} \ \int\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\text{dx} \ \ ...(1)$
Put $\frac{2\text{x}^2+\text{x}}{\text{(x}-1)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1} \ \ ...(2 )$
$\Rightarrow \ 2\text{x}^2 +\text{x}=\text{Ax}^2 + \text{A} +\text{Bx}^2 +\text{Cx + Bx + C}$
$\Rightarrow \ \ 2\text{x}^2 +\text{x} = (\text{A + B)x}^2+\ (\text{B + C)x + A + C} \ \ \ \ ......(3)$
Now comparing the coefficients of $x^2$ and $x$
$\Rightarrow \text{A + B} =2$
$\Rightarrow \text{B + C}=1$
$\Rightarrow \text{A + C}=0{}$
Solving tham we will get the values of $A, B, C$
$\text{A}= \frac{1}{2}, \text{B} =\frac{3}{2}-\frac{1}{2}$
Putting the values of $A,B,C$
$\therefore \text{from}(2), \frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)} = \frac{\frac{1}{2}}{\text{x}+1}+\frac{\frac{3}{2}\text{x}-\frac{1}{2}}{\text{x}^2+1}$
$\text{or}\ \ \frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)} \equiv \frac{1}{2(\text{x}+1)}+\frac{3}{2}\bigg(\frac{x}{x^2+1}\bigg)-\frac{1}{2}\bigg(\frac{1}{x^2+1}\bigg)$
$\therefore \text{from}(1), \ \int\text{dy}=\frac{1}{2} \int\frac{1}{x+1}\text{dx}+\frac{3}{2}\int\frac{\text{x}}{\text{x}^2+1}\text{dx}-\frac{1}{2}\int \frac{1}{\text{x}^2+1}\text{dx}$
$\therefore \ \int 1\ \text{dy}= \frac{1}{2}\int\frac{1}{\text{x}+1}\text{dx}+\frac{3}{4}\int\frac{2\text{x}}{\text{x}^2+1} \text{dx}-\frac{1}{2}\int\frac{1}{\text{x}^2+1}\text{dx}$
$\therefore \ \text{y}= \frac{1}{2}\text{log}|\text{x}+1|+ \frac{3}{4}\text{log}(\text{x}^2+1)-\frac{1}{2}\text{tan}^{-1}\text{x}+\text{c} \ \ \ \ \ .....(4)$
Now $y = 1$ when $x = 0$
$\therefore 1=\frac{1}{2}\text{log}(1)+\frac{3}{4}\text{log} \ 1-\frac{1}{2}\text{tan}^{-1} \text{0+c}$
$\therefore 1 =\frac{1}{2}(0)+\frac{3}{4}(0)-\frac{1}{2}(0)+\text{c} \ \ \Rightarrow\ \ \text{c}=1$
$\therefore \text{from}\ (4),\ \text{y}= \frac{1}{2}\text{log}|\text{x}+1|+\frac{3}{4}\text{log}(\text{x}^2+1)-\frac{1}{2}\text{tan}^{-1}\text{x}+1$

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