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Beliebt Rechnen >

y^{''}=y^'+y-1,y(0)=0,y^'(0)=0

Lösung

y^{^{''}} = y^{^{'}} + y - 1, y (0) = 0, y^{^{'}} (0) = 0

Lösung

y = (\frac{1 + 5}{2 5} - 1) e^{\frac{( 1 + 5 ) t}{2}} - \frac{1 + 5}{2 5} e^{\frac{( 1 - 5 ) t}{2}} + 1

Schritte zur Lösung

y^{''} (t) = y^{'} (t) + y - 1

Solve linear ODE:

y = (\frac{1 + 5}{2 5} - 1) e^{\frac{( 1 + 5 ) t}{2}} - \frac{1 + 5}{2 5} e^{\frac{( 1 - 5 ) t}{2}} + 1

y = (\frac{1 + 5}{2 5} - 1) e^{\frac{( 1 + 5 ) t}{2}} - \frac{1 + 5}{2 5} e^{\frac{( 1 - 5 ) t}{2}} + 1

Graph

Beliebte Beispiele

2y^{''}+3y^'+8y=g(x)

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

(d^2-2d+1)y=xe^xsin(x)

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

y^{^{''}} + 4 y = 3 t + 2

y^{''}-2y^'+y=(4x^2-3+e^x)/x

y^{^{''}} - 2 y^{^{'}} + y = \frac{4 x ^{2} - 3 + e ^{x}}{x}

y^{'''}+y^{''}=x^2

y^{^{'''}} + y^{^{''}} = x^{2}