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Beliebt Rechnen >

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

Lösung

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

Lösung

y = cos (t) + sin (t) + 1

Schritte zur Lösung

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

Solve linear ODE:

y = cos (t) + sin (t) + 1

y = cos (t) + sin (t) + 1

Graph

Beliebte Beispiele

y^{''}-2y^'+y=t^{-4}e^t

y^{^{''}} - 2 y^{^{'}} + y = t^{- 4} e^{t}

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

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

(d^2y)/(dx^2)-6y=e^{3x}(x^2+2)

\frac{d ^{2} y}{d x ^{2}} - 6 y = e^{3 x} (x^{2} + 2)

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

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

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

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