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Beliebt Rechnen >

y^{''}+y=cos^2(t),y(0)=0,y^'(0)=0

Lösung

y^{^{''}} + y = cos^{2} (t), y (0) = 0, y^{^{'}} (0) = 0

Lösung

y = \frac{1}{3} cos^{4} (t) + sin (t) (sin (t) - \frac{sin ^{3} ( t )}{3}) - \frac{1}{3} cos (t)

Schritte zur Lösung

y^{''} (t) + y = cos^{2} (t)

Solve linear ODE:

y = \frac{1}{3} cos^{4} (t) + sin (t) (sin (t) - \frac{sin ^{3} ( t )}{3}) - \frac{1}{3} cos (t)

y = \frac{1}{3} cos^{4} (t) + sin (t) (sin (t) - \frac{sin ^{3} ( t )}{3}) - \frac{1}{3} cos (t)

Graph

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