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Beliebt Rechnen >

y^{''}+4y=t^2+2e^t,y(0)=0,y^'(0)=1

Lösung

y^{^{''}} + 4 y = t^{2} + 2 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

Lösung

y = - \frac{11}{40} cos (2 t) + \frac{3}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{2}{5} e^{t}

Schritte zur Lösung

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

Solve linear ODE:

y = - \frac{11}{40} cos (2 t) + \frac{3}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{2}{5} e^{t}

y = - \frac{11}{40} cos (2 t) + \frac{3}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{2}{5} e^{t}

Graph

Beliebte Beispiele

y^{''}-10y^'+25y=30x+12

y^{^{''}} - 10 y^{^{'}} + 25 y = 30 x + 12

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

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

y^{''}+2y=sin(x)

y^{^{''}} + 2 y = sin (x)

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

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

y^{''}-4y^'+8y=e^xcos(2x)

y^{^{''}} - 4 y^{^{'}} + 8 y = e^{x} cos (2 x)