Solve the differential equation left( {1 + xs | vedclass

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(B) Given the differential equation: $\left( {1 + x\sqrt {{x^2} + {y^2}} } \right)dx + \left( {\sqrt {{x^2} + {y^2}} - 1 } \right)y dy = 0$. Rearrangi

Vedclass · Accepted Answer

$x - \frac{{{y^2}}}{2} + \frac{1}{3}{\left( {{x^2} + {y^2}} \right)^{\frac{3}{2}}} = C$

Vedclass · Answer

(B) Given the differential equation: $\left( {1 + x\sqrt {{x^2} + {y^2}} } \right)dx + \left( {\sqrt {{x^2} + {y^2}} - 1 } \right)y dy = 0$. Rearranging the terms,we get: $dx - y dy + \sqrt{x^2 + y^2} (x dx + y dy) = 0$. We know that $d(x^2 + y^2) = 2x dx + 2y dy$,so $x dx + y dy = \frac{1}{2} d(x^2 + y^2)$. Substituting this into the equation: $dx - y dy + \frac{1}{2} \sqrt{x^2 + y^2} d(x^2 + y^2) = 0$. Integrating both sides: $\int dx - \int y dy + \frac{1}{2} \int (x^2 + y^2)^{1/2} d(x^2 + y^2) = C$. This gives: $x - \frac{y^2}{2} + \frac{1}{2} \cdot \frac{(x^2 + y^2)^{3/2}}{3/2} = C$. Simplifying,we get: $x - \frac{y^2}{2} + \frac{1}{3} (x^2 + y^2)^{3/2} = C$.

Solve the differential equation $\left( {1 + x\sqrt {{x^2} + {y^2}} } \right)\,dx + \left( {\sqrt {{x^2} + {y^2}} - 1 } \right)y\,dy = 0$.

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