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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation
$\text{x}(\text{x}^{2} - 1)\frac{\text{dy}}{\text{dx}} = 1, \text{y}(2) = 0$
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Answer

We have
$\text{x}(\text{x}^{2} - 1)\frac{\text{dy}}{\text{dx}} = 1$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{1}{\text{x}(\text{x}^{2} - 1)}$
$\Rightarrow\text{dy} = \bigg\{\frac{1}{\text{x}(\text{x}^{2} - 1)}\bigg\} \text{dx}$
Integrating both sides, we get
$\int \text{dy} = \int \bigg\{\frac{1}{\text{x}(\text{x}^{2} - 1)}\bigg\} \text{dx}$
$\Rightarrow \text{y} = \int\bigg\{\frac{1}{\text{x}(\text{x}^{2} - 1)}\bigg\} \text{dx}$
$\Rightarrow \text{y} = \int \frac{1}{\text{x}(\text{x} - 1)(\text{x} + 1)} \text{dx}$
$$Let $\frac{1}{\text{x}(\text{x} - 1)(\text{x} + 1)} = \frac{\text{A}}{\text{x}} + \frac{\text{B}}{\text{x}-1} + \frac{\text{C}}{\text{x}+ 1}$
$\Rightarrow 1 = \text{A} (\text{x}^{2} - 1)+\text{B} (\text{x}^{2} +\text{x) } + \text{C} (\text{x}^{2} - \text{x})$
$\Rightarrow 1 = (\text{A + B + C}) \text{x}^{2} + (\text{B} - \text{C}) \text{x} - \text{A}$
Equating the coefficients on both sides we get
A + B + C = 0 .....(1)
B - C = 0 .....(2)
A = -1 .....(3)
Solving (1), (2) and (3), we get
$\text{A} = -1$
$\text{B} = \frac{1}{2}$
$\text{C} = \frac{1}{2}$
$\therefore \text{y} = \frac{1}{2} \int \frac{1}{\text{x} - 1} \text{dx} - \int \frac{1}{\text{x}} \text{dx} + \frac{1}{2} \int \frac{1}{\text{x + 1}} \text{dx}$
$= \frac{1}{2} \log |\text{x } - 1| - \log|\text{x}| + \frac{1}{2}\log|\text{x + 1}| + \text{C}$
$= \frac{1}{2} \log |\text{x} - 1| + \frac{1}{2} \log | \text{x + 1} | - \log |\text{x}| + \text{C}$
It is given that y(2) = 0.
$\therefore 0 = \frac{1}{2} \log |2 - 1| + \frac{1}{2}\log|2+1|-\log|2|+\text{C}$
$\text{C}=\log|2|-\frac{1}{2}-\frac{1}{2}\log|3|$
Substituting the value of C, we get
$\text{y}=\frac{1}{2}\log|\text{x}-1|+\frac{1}{2}\log|\text{x}+1|-\log|\text{x}|+\log|2|-\frac{1}{2}\log|3|$
$\Rightarrow2\text{y}=\log|\text{x}-1|+\log|\text{x}+1|-2\log|\text{x}|+2\log|2|-\log|3|$
$\Rightarrow2\text{y}=\log|\text{x}-1+\log|\text{x}+1|-\log|\text{x}^3|+\log4-\log3$
$\Rightarrow2\text{y}=\log\frac{4(\text{x}-1)(\text{x}+1)}{3\text{x}^2}$
$\Rightarrow\text{y}=\frac{1}{2}\log\frac{4(\text{x}^2-1)}{3\text{x}^2}$
Hence, $\text{y}=\frac{1}{2}\log\frac{4(\text{x}^2-1)}{3\text{x}^2}$ is the solution to the given differerntial equation.

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