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

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

Answers (1)

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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