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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 45 maths textbook solution.

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Answer : y=c e^{-2 \tan ^{-1} x}+\frac{1}{2}

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

Given:  \left(x^{2}+1\right) d y+(2 y-1) d x=0

Solution : \left(x^{2}+1\right) d y+(2 y-1) d x=0

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

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

Comparing with \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{p} \mathrm{y}=\mathrm{q}, we get,

\begin{aligned} &\mathrm{P}=\frac{2}{\left(1+\mathrm{x}^{2}\right)} \quad \text { and } \quad \mathrm{q}=\frac{1}{1+\mathrm{x}^{2}} \\ &\text { Now, I.F }=\mathrm{e}^{\int \mathrm{p} \mathrm{dx}} \end{aligned}

\begin{aligned} &=e^{\int 2 \frac{d x}{1+x^{2}}} \\ &=e^{2 \tan ^{-1} x} \end{aligned}

The solution is,

\begin{aligned} &\mathrm{y} \mathrm{x} \mathrm{I.F}=\int \mathrm{Q} \times \mathrm{I} \cdot \mathrm{F}+\mathrm{C} \\ &\mathrm{y} \mathrm{e}^{2 \tan ^{-1} \mathrm{x}}=\int \frac{\mathrm{e}^{2 \tan ^{-1} \mathrm{x}}}{1+\mathrm{x}^{2}} \mathrm{dx} \end{aligned}

Put \tan^{-1}x=t, differentiating both side,

\begin{aligned} &\frac{1}{1+x^{2}} d x=d t \\ &y_{n} e^{2 \tan ^{-1} x}=\int e^{2 t} d t \\ &y e^{2 \tan ^{-1} x}=\frac{1}{2} e^{2 t}+c_{m}\left[\int e^{a t} d x=\frac{1}{a} e^{a x}+c\right] \end{aligned}

\begin{aligned} &\text { y } e^{2 \tan ^{-1} x}=\frac{1}{2} e^{2 \tan ^{-1} x}+c \\ &y=c e^{-2 \tan ^{-1} x}+\frac{1}{2} \end{aligned}

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