y^'=((2y+3)/(4t+5))^2

Lösung

y^{^{'}} = (\frac{2 y + 3}{4 t + 5})^{2}

y = - \frac{4 t}{- 1 + 16 c _{1} t + 20 c _{1}} - \frac{24 c _{1} t}{- 1 + 16 c _{1} t + 20 c _{1}} - \frac{30 c _{1}}{- 1 + 16 c _{1} t + 20 c _{1}} - \frac{7}{2 ( - 1 + 16 c _{1} t + 20 c _{1} )}

Schritte zur Lösung

y^{'} (t) = (\frac{2 y + 3}{4 t + 5})^{2}

Solve separable ODE:

y = - \frac{4 t}{- 1 + 16 c _{1} t + 20 c _{1}} - \frac{24 c _{1} t}{- 1 + 16 c _{1} t + 20 c _{1}} - \frac{30 c _{1}}{- 1 + 16 c _{1} t + 20 c _{1}} - \frac{7}{2 ( - 1 + 16 c _{1} t + 20 c _{1} )}

y = - \frac{4 t}{- 1 + 16 c _{1} t + 20 c _{1}} - \frac{24 c _{1} t}{- 1 + 16 c _{1} t + 20 c _{1}} - \frac{30 c _{1}}{- 1 + 16 c _{1} t + 20 c _{1}} - \frac{7}{2 ( - 1 + 16 c _{1} t + 20 c _{1} )}

x \frac{d y}{d x} = e^{y + x}, x > 0

y^{^{'}} - \frac{3 y}{x} = 0

(1 - x) y^{^{'}} = 6 y

x y^{2} d y + (x^{2} + 1) d x = 0

\frac{d y}{x - 1} = y d x