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  • Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 49 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 49 maths textbook solution.

Answers (1)

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

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

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

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

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

putting y=vx and differentiate,

      \begin{aligned} &\frac{d y}{d x}=v+x \frac{d v}{d x} \\ &\left.v+x \frac{d v}{d x}=\frac{x y-x^{2}-y^{2}}{x^{2}} \quad \text { [put value of } \frac{d y}{d x}\right] \\ &v+x \frac{d v}{d x}=v-1-v^{2} \end{aligned}

      \begin{aligned} &\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}-1-\mathrm{v}^{2}-\mathrm{v} \\ &\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-1-\mathrm{v}^{2} \\ &\frac{\mathrm{dv}}{1+\mathrm{v}^{2}}=-\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}

Integrating both sides,

    \begin{aligned} &\int \frac{d v}{1+v^{2}} d v=-\int \frac{1}{x} d x \\ &\tan ^{-1} v=-\log |x|+\log c\left[\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x\right] \\ &\tan ^{-1} v=\log \frac{c}{x} \end{aligned}

    \begin{aligned} &\mathrm{e}^{\tan ^{-1} v}=\frac{c}{\mathrm{x}}\left[\log \mathrm{x}=\mathrm{a}, \mathrm{x}=\mathrm{e}^{\mathrm{a}}\right]\\ &\left.c=x e^{\tan ^{-1} \frac{y}{x}} \text { [put value of } v=\frac{x}{y}\right] \end{aligned}

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