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Beliebt Rechnen >

y^{''}+4y=t^2+6e^t,y(0)=0,y^'(0)=3

Lösung

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 3

Lösung

y = - \frac{43}{40} cos (2 t) + \frac{9}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{6}{5} e^{t}

Schritte zur Lösung

y^{''} (t) + 4 y = t^{2} + 6 e^{t}

Solve linear ODE:

y = - \frac{43}{40} cos (2 t) + \frac{9}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{6}{5} e^{t}

y = - \frac{43}{40} cos (2 t) + \frac{9}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{6}{5} e^{t}

Graph

Beliebte Beispiele

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

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

y^{'''}-y^'=e^{5t}

y^{^{'''}} - y^{^{'}} = e^{5 t}

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

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

y^{''}+9y=e^{2t}\H(t-1)

y^{^{''}} + 9 y = e^{2 t} H (t - 1)

y^{''}+2y^'+y=1,y(0)=2

y^{^{''}} + 2 y^{^{'}} + y = 1, y (0) = 2