y^{''}-0.2y^'+0.01y=0,y(0)=12,y^'(0)=b

Lösung

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 12, y^{^{'}} (0) = b

y = 12 e^{\frac{t}{10}} + (b - \frac{6}{5}) e^{\frac{t}{10}} t

Schritte zur Lösung

y^{''} (t) - 0.2 y^{'} (t) + 0.01 y = 0

Solve linear ODE:

y = 12 e^{\frac{t}{10}} + (b - \frac{6}{5}) e^{\frac{t}{10}} t

y = 12 e^{\frac{t}{10}} + (b - \frac{6}{5}) e^{\frac{t}{10}} t

\frac{d}{d x} ((\frac{x ^{2}}{8} + x - \frac{1}{x})^{4})

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in v erse l a pl a ce \frac{( 2 s + 3 )}{s ^{2} + 6 s + 13}