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  • Explain solution RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 21 Maths.

Explain solution RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 21 Maths.

Answers (1)

Answer: y=\ln \left |x+1 \right |+\frac{1}{2}\ln \left |x^2+1 \right |-\int \frac{dx}{(x^2+1)(x+1)}

Hint: You have to integrate by applying integration of xn.

Given: (x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x

Solution: (x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x

\begin{aligned} &{\left[x^{2}(x+1)+1(x+1)\right] \frac{d y}{d x}=2 x^{2}+x} \\ &{\left[\left(x^{2}+1\right)(x+1)\right] \frac{d y}{d x}=2 x^{2}+x} \\ &d y=\frac{x(2 x+1)}{\left(x^{2}+1\right)(x+1)} d x \end{aligned}

Integrate on both sides

  \int d y=\int \frac{x(2 x+1)}{\left(x^{2}+1\right)(x+1)} d x            ..........…(i)

By partial fractions,

\begin{aligned} &\frac{2 x^{2}+x}{\left(x^{2}+1\right)+(x+1)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+1} \\ &2 x^{2}+x=A\left(x^{2}+1\right)+(B x+C)(x+1) \\ &2 x^{2}+x=x^{2}(A+B)+x(B+C)+(A+C) \end{aligned}

Compare on both sides

A+B=2,B+C=1,A+C=0

\begin{aligned} &B=\frac{3}{2^{\prime}} A=\frac{1}{2^{\prime}} C=-\frac{1}{2} \\ &=y=\int\left[\frac{\frac{1}{2}}{x+1}+\frac{\frac{3}{2} x-\frac{1}{2}}{x^{2}+1}\right] d x \\ &=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{3}{2} \int \frac{x}{x^{2}+1} d x-\frac{1}{2} \int \frac{1}{x^{2}+1} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{1}{x+1}=\log |x+1|, \frac{1}{x^{2}+1}=\tan ^{-1} x\right] \\ &y=\frac{1}{2} \log |x+1|+\frac{3}{4} \log \left|x^{2}+1\right|-\frac{1}{2} \tan ^{-1} x+C \end{aligned}

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