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Differential Equations — MATHS STD 12 Science — Question

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

Answer

We have,
$(\text{x}^2-1)\frac{\text{dy}}{\text{dx}}+2(\text{x}+2)\text{y}=2(\text{x}+1)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{2(\text{x}+2)}{\text{x}^2-1}\text{y}=\frac{2(\text{x}+1)}{\text{x}^2-1}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{2(\text{x}+2)}{\text{x}^2-1}\text{y}=\frac{2}{\text{x}-1}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\frac{2(\text{x}+2)}{\text{x}^2-1}$
$\text{Q}=\frac{2}{\text{x}-1}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\frac{2(\text{x+2})}{\text{x}^2-1}\text{dx}}$
$=\text{e}^{\int\frac{2\text{x}}{\text{x}^2-1}+4\int\frac{1}{\text{x}^2-1}\text{dx}}$
$=\text{e}^{\log|\text{x}^2-1|+4\times\frac{1}2\log\big|\frac{\text{x}-1}{\text{x}+1}\big|}$
$=\text{e}^{\log\Big|(\text{x}^2-1)\times\frac{(\text{x}-1)^2}{(\text{x}+1)^2}\Big|}$
$=\text{e}^{\log\Big|(\text{x}^2-1)\times\frac{(\text{x}-1)^2}{(\text{x}+1)^2}\Big|}$
$=\text{e}^{\log\Big|\frac{(\text{x}-1)^3}{(\text{x}+1)}\Big|}$
$=\frac{(\text{x}-1)^3}{(\text{x}+1)}$
Multiplying both sides of (1) by $\frac{(\text{x}-1)^3}{(\text{x}+1)},$ we get
$\frac{(\text{x}-1)^3}{(\text{x}+1)}\Big(\frac{\text{dy}}{\text{dx}}+\frac{2(\text{x}+2)}{\text{x}^2-1}\text{y}\Big)=\frac{(\text{x}-1)^3}{(\text{x}+1)}\times\frac{2}{\text{x}-1}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\frac{\text{dy}}{\text{dx}}-\frac{2(\text{x}+2)(\text{x}-1)^2}{(\text{x}+1)^2}\text{y}=\frac{2(\text{x}-1)^2}{(\text{x}+1)}$
Integrating both sides with respect to x, we get
$\frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int\frac{2(\text{x}-1)^2}{(\text{x}+1)}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int2\Big\{\frac{(\text{x}-1)^2-4\text{x}}{(\text{x}+1)}\Big\}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int2\Big\{(\text{x}+1)-\frac{4\text{x}}{(\text{x}+1)}\Big\}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int2\Big\{(\text{x}+1)-\frac{4(\text{x}+1-1)}{(\text{x}+1)}\Big\}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int2\Big\{(\text{x}+1)-4+\frac{4}{(\text{x}+1)}\Big\}\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\int\Big(2\text{x}-6+\frac{8}{\text{x}+1}\Big)\text{dx + C}$
$\Rightarrow\ \frac{(\text{x}-1)^3}{(\text{x}+1)}\text{y}=\text{x}^2-6\text{x}+8\log|\text{x}+1|+\text{C}$
$\Rightarrow\ \text{y}=\frac{(\text{x}-1)^3}{(\text{x}+1)}(\text{x}^2-6\text{x}+8\log|\text{x}+1|+\text{C})$
Hence, $\text{y}=\frac{(\text{x}-1)^3}{(\text{x}+1)}(\text{x}^2-6\text{x}+8\log|\text{x}+1|+\text{C})$ is the required solution.

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