y^'=((x^4+x^2y^2+y^4))/(x^3y)

Lösung

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

y = tan (2 ln (x) + c_{1}) x, y = - x tan (2 ln (x) + c_{1})

Schritte zur Lösung

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

Solve using substitution:

y = tan (2 ln (x) + c_{1}) x, y = - x tan (2 ln (x) + c_{1})

y = tan (2 ln (x) + c_{1}) x, y = - x tan (2 ln (x) + c_{1})

l a pl a ce t r an s f or m t - 2

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