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Differential Equations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
Solve the following differential equations:
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{x}^3\text{y}$
✓

Answer

We have,
$(\text{x}-1)\frac{\text{dy}}{\text{dx}}=2\text{x}^3\text{y}$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\frac{2\text{x}^3}{\text{x}-1}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\frac{2\text{x}^3}{\text{x}-1}\text{dx}$
$\Rightarrow\log|\text{y}|=2\int\frac{\text{x}^3-1+1}{\text{x}-1}\text{dx}$
$\Rightarrow\log|\text{y}|=2\Big[\int\frac{\text{x}^3-1}{\text{x}-1}\text{dx}+\int\frac{1}{\text{x}-1}\text{dx}\Big]$
$\Rightarrow\log|\text{y}|=2\Big[\int\frac{(\text{x}-1)(\text{x}^2+\text{x}+1)}{\text{x}-1}\text{dx}+\int\frac{1}{\text{x}-1}\text{dx}\Big]$
$\Rightarrow\log|\text{y}|=2\Big[\int(\text{x}^2+\text{x}+1)\text{dx}+\int\frac{1}{\text{x}-1}\text{dx}\Big]$
$\Rightarrow\log|\text{y}|=2\Big[\frac{\text{x}^3}{3}+\frac{\text{x}^2}{2}+\text{x}+\log|\text{x}-1|\Big]+\text{C}$
$\Rightarrow\log|\text{y}|=\frac{2}{3}\text{x}^3+\text{x}^2+2\text{x}+\log|\text{x}-1|^2+\text{C}$
$\Rightarrow\text{y}=\text{e}^{\frac{2}{3}}\text{x}^3+\text{x}^2+2\text{x}+\log|\text{x}-1|^2+\text{C}$
$\Rightarrow\text{y}=\text{e}^{\text{C}}\times\text{e}^{\log|\text{x}-1|^2}\times\text{e}^{\frac{2}{3}}\text{x}^3+\text{x}^2+2\text{x}$
$\Rightarrow\text{y = C}_1|\text{x}-1|^2\text{e}^{\frac{2}{3}}\text{x}^3+\text{x}^2+2\text{x}$ $\big[\because\text{e}^{\text{In x}}=\text{x}\text{and where, C}_1=\text{e}^{\text{c}}\big]$
$\therefore\text{y}=\text{C}_1|\text{x}-1|^2\text{e}^{\frac{2}{3}}\text{x}^3+\text{x}^2+2\text{x}$ is required solution.

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