y^'=x^2y^22^x

Lösung

y^{^{'}} = x^{2} y^{2} 2^{x}

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

Schritte zur Lösung

y^{'} (x) = x^{2} y^{2} 2^{x}

Solve separable ODE:

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

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

y^{^{'}} = a \cdot y

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

y^{^{'}} = e^{x^{2}} 100 - y^{2}

x^{^{'}} = \frac{x}{t}

\frac{d y}{d x} = - \frac{7}{8} x^{3} y^{7}