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

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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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