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

Lösung

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

y = 1 + t^{3} + 2 t^{2} + 2 t + c_{1} + 1, y = 1 - t^{3} + 2 t^{2} + 2 t + c_{1} + 1

Schritte zur Lösung

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

Solve separable ODE:

y = 1 + t^{3} + 2 t^{2} + 2 t + c_{1} + 1, y = 1 - t^{3} + 2 t^{2} + 2 t + c_{1} + 1

y = 1 + t^{3} + 2 t^{2} + 2 t + c_{1} + 1, y = 1 - t^{3} + 2 t^{2} + 2 t + c_{1} + 1