y+y^4=(y^6+3x)y^'

Lösung

y + y^{4} = (y^{6} + 3 x) y^{^{'}}

y = - 3 \frac{3 c _{1} - 3 x + ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}, y = - 3 \frac{3 c _{1} - 3 x - ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}

Schritte zur Lösung

y + y^{4} = (y^{6} + 3 x) y^{'} (x)

Solve exact ODE:

y = - 3 \frac{3 c _{1} - 3 x + ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}, y = - 3 \frac{3 c _{1} - 3 x - ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}

y = - 3 \frac{3 c _{1} - 3 x + ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}, y = - 3 \frac{3 c _{1} - 3 x - ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}

cos (x) y^{^{'}} - sin (x) y = 0

\int \frac{9 x}{4 x + 2} d x

\int \frac{3 - z}{- 1 + z ^{2}} d z

d er i v a t i v e cos (sin (5) x^{3})

\frac{d}{d x} (ln (\frac{x}{a}))