y^'+8y=3x

Lösung

y^{^{'}} + 8 y = 3 x

y = \frac{3 ( 8 x - 1 )}{64} + \frac{c _{1}}{e ^{8 x}}

Schritte zur Lösung

y^{'} (x) + 8 y = 3 x

Solve linear ODE:

y = \frac{3 ( 8 x - 1 )}{64} + \frac{c _{1}}{e ^{8 x}}

y = \frac{3 ( 8 x - 1 )}{64} + \frac{c _{1}}{e ^{8 x}}

\frac{\partial}{\partial u} (2 uv)

d er i v a t i v e (x - 2) (3 x + 1)^{2}

\frac{d y}{d x} = e^{6 x} + 9 y

a re a 5 x - x^{2}, x, 4, 4

\int cos^{3} (y) d y