dy=θ^3dθ

Lösung

d y = θ^{3} d θ

θ = 4 4 y + c_{1}, θ = - 4 4 y + c_{1}

Schritte zur Lösung

d y = θ^{3} d θ

Solve separable ODE:

θ = 4 4 y + c_{1}, θ = - 4 4 y + c_{1}

θ = 4 4 y + c_{1}, θ = - 4 4 y + c_{1}

\frac{d y}{d x} y = \frac{( 5 x ^{2} + 2 ) ( 3 x + 9 )}{8 x - 9}

3 e^{2 x} \frac{d y}{d x} = - 4 \cdot \frac{x}{y ^{2}}

\frac{d y}{d x} = \frac{( 1 - y )}{xv}

y^{^{'}} = 1 - \frac{y}{10}

20 \frac{d y}{d x} + y = 1