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Beliebt Rechnen >

a^2y+y^{''}=cos(bt),y(0)=0,y^'(0)=0

Lösung

a^{2} \cdot y + y^{^{''}} = cos (b \cdot t), y (0) = 0, y^{^{'}} (0) = 0

Lösung

y = \frac{cos ( b t )}{a ^{2} - b ^{2}} - \frac{cos ( a t )}{a ^{2} - b ^{2}}

Schritte zur Lösung

a^{2} \cdot y + y^{''} (t) = cos (b \cdot t)

Solve linear ODE:

y = \frac{cos ( b t )}{a ^{2} - b ^{2}} - \frac{cos ( a t )}{a ^{2} - b ^{2}}

y = \frac{cos ( b t )}{a ^{2} - b ^{2}} - \frac{cos ( a t )}{a ^{2} - b ^{2}}

Graph

Beliebte Beispiele

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

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

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

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

y^{''}+8y^'+16y=(3e^{-4x})/x

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x}

(d^3-3d^2-10d+24)y=2x^2+5

(d^{3} - 3 d^{2} - 10 d + 24) y = 2 x^{2} + 5

linear y^{''}-y^'+2=0

l in e a r y^{^{''}} - y^{^{'}} + 2 = 0