#### Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 50 maths textbook solution.

Answer : $x=c^{1}(y-b)(b x+1)$

Hint: you must know the rules of solving differential equation and integrations.

Given: $(y-x) \frac{d y}{d x}=b\left(1+x^{2} \frac{d y}{d x}\right)$

Solution : $(y-x) \frac{d y}{d x}=b\left(1+x^{2} \frac{d y}{d x}\right)$

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

Integrating both sides,

\begin{aligned} &\int \frac{1}{y-b} d y=\int \frac{1}{b x^{2}+x} d x \\ &\int \frac{1}{y-b} d y=\frac{1}{b} \int \frac{1}{x^{2}+\frac{x}{b}} d x \\ &\int \frac{1}{y-b} d y=\frac{1}{b} \int\left(\frac{1}{x^{2}+\frac{x}{b}+\frac{1}{4 b^{2}}-\frac{1}{4 b^{2}}}\right) d x \end{aligned}

$\int\left(\frac{1}{y-b}\right) d y=\frac{1}{b} \int\left(\frac{1}{\left(x+\frac{1}{2 b}\right)^{2}-\left(\frac{1}{2 b}\right)^{2}}\right) d x$

\begin{aligned} &\Rightarrow \log |y-b|=\frac{1}{2 \operatorname{x} \frac{1}{2 \mathrm{~b}} \times \mathrm{b}} \log \left|\frac{x+\frac{1}{2 b}-\frac{1}{2 b}}{x+\frac{1}{2 b}+\frac{1}{2 b}}\right|+\log c\left[\because \int \frac{1}{\mathrm{x}} \mathrm{dx}=\log |\mathrm{x}| \text { and } \int \frac{1}{(\mathrm{x}+\mathrm{a})^{2}-\mathrm{a}^{2}}=\log \left|\frac{x+a-a}{x+a+a}\right|\right] \\ &\Rightarrow \log |y-b|=\log \left|\frac{b x}{b x+1}\right|+\log \mathrm{c} \end{aligned}

\begin{aligned} & \Rightarrow \quad y-b=\frac{c b x}{b x+1} \\ & \Rightarrow \quad c b x=(y-b)(b x+1) \\ \therefore x &=c^{1}(y-b)(b x+1) \quad\left[\text { where } \frac{1}{c b}=c^{1}\right] \end{aligned}