x^3y^'+2x^2y=-2cos(2x)y^{-1/2}

Lösung

x^{3} y^{^{'}} + 2 x^{2} y = - 2 cos (2 x) y^{- \frac{1}{2}}

y = \frac{( 9 sin ^{2} ( 2 x ) + 4 ( - 3 c _{1} sin ( 2 x ) + c _{1}^{2} ) ) ^{\frac{1}{3}}}{4 ^{\frac{1}{3}} x ^{2}}

Schritte zur Lösung

x^{3} y^{'} (x) + 2 x^{2} y = - 2 cos (2 x) y^{- \frac{1}{2}}

Solve Bernoulli ODE:

y = \frac{( 9 sin ^{2} ( 2 x ) + 4 ( - 3 c _{1} sin ( 2 x ) + c _{1}^{2} ) ) ^{\frac{1}{3}}}{4 ^{\frac{1}{3}} x ^{2}}

y = \frac{( 9 sin ^{2} ( 2 x ) + 4 ( - 3 c _{1} sin ( 2 x ) + c _{1}^{2} ) ) ^{\frac{1}{3}}}{4 ^{\frac{1}{3}} x ^{2}}

d er i v a t i v e f (x) = (x^{2} + 3 x) (x - 1)

\frac{d}{d x} (120 x^{3} - 180 x^{2} + 60)

in v erse l a pl a ce \frac{6 e ^{- s}}{s ( 10 s + 1 )}

\frac{\partial}{\partial x} (\frac{( 3 )}{( x + y ) ^{3}})

\int sec^{3} (x) tan (x) d x