y^{''}-2y^'+1=e^x

Lösung

y^{^{''}} - 2 y^{^{'}} + 1 = e^{x}

y = c_{1} e^{2 x} + c_{2} - e^{x} + \frac{x}{2}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - e^{x} + \frac{x}{2}

y = c_{1} e^{2 x} + c_{2} - e^{x} + \frac{x}{2}

y^{^{''}} - 4 y = cos^{2} (x)

y^{^{''}} + 9 y = 5 csc^{2} (3 x)

y^{^{''}} - y - 12 y = e^{4 x}

\frac{d ^{2} y}{d x ^{2}} - y = \frac{1}{x}

6 y^{^{''}} - y^{^{'}} - 2 y = 3 sin (10) x