Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
Solve the following differential equation $(\text{x}^3+\text{x}^2+\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{x}^2+\text{x}$
✓
Answer
We have,
$(\text{x}^3+\text{x}^2+\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{x}^2+\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^2+\text{x}}{\text{x}^3+\text{x}^2+\text{x}+1}$
$\Rightarrow\text{dy}=\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\ \text{dx}$
Intregrating both sides, we get
$\int\text{dy}=\int\Big\{\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\Big\}\text{dx}$
$\Rightarrow\text{y}=\int\Big\{\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}\Big\}\text{dx}$
Let $\frac{2\text{x}^2+\text{x}}{(\text{x}+1)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow 2x^2 + x = Ax^2 + A + Bx^2 + Bx + Cx + C$
$\Rightarrow 2x^2 + x = (A + B)x^{2 }+ (B + C)x + (A + C)$
Compairing the coefficient on both sides, we get
$A + B = 2 ...(1)$
$B + C = 1 ...(2)$
$A + C = 0 ...(3)$
Solving $(1), (2)$ and $(3),$ we get
$\text{A}=\frac{1}{2}$
$\text{B}=\frac{3}{2}$
$\text{C}=-\frac{1}{2}$
$\therefore\text{y}=\frac{1}{2}\int\frac{1}{(\text{x}+1)}\ \text{dx}+\int\frac{\frac{3}{2}\text{x}-\frac{1}{2}}{\text{x}^2+1}\ \text{dx}$
$=\frac{1}{2}\int\frac{1}{(\text{x}+1)}\text{dx}+\frac{3}{4}\int\frac{2\text{x}}{\text{x}^2+1}\ \text{dx}-\frac{1}{2}\int\frac{1}{\text{x}^2+1}\ \text{dx}$
$=\frac{1}{2}\log|\text{x}+1|+\frac{3}{4}\log|\text{x}^2+1|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
Hence, $\text{y}=\frac{1}{2}\log|\text{x}+1|+\frac{3}{4}\log|\text{x}^2+1|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$ is the solution to the given differential equation.
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