4xyy^'=4y^2+3xsqrt(x^2+y^2)

Lösung

4 x y y^{^{'}} = 4 y^{2} + 3 x x^{2} + y^{2}

y = \frac{9 ln ^{2} ( x ) + 24 c _{1} ln ( x ) + 16 c _{1}^{2} - 16 x}{4}, y = - \frac{x 9 ln ^{2} ( x ) + 24 c _{1} ln ( x ) + 16 c _{1}^{2} - 16}{4}

Schritte zur Lösung

4 x y y^{'} (x) = 4 y^{2} + 3 x x^{2} + y^{2}

Solve using substitution:

y = \frac{9 ln ^{2} ( x ) + 24 c _{1} ln ( x ) + 16 c _{1}^{2} - 16 x}{4}, y = - \frac{x 9 ln ^{2} ( x ) + 24 c _{1} ln ( x ) + 16 c _{1}^{2} - 16}{4}

y = \frac{9 ln ^{2} ( x ) + 24 c _{1} ln ( x ) + 16 c _{1}^{2} - 16 x}{4}, y = - \frac{x 9 ln ^{2} ( x ) + 24 c _{1} ln ( x ) + 16 c _{1}^{2} - 16}{4}

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