y''+5y'+2y=e^(-t)cos(2t), y(0)=6, y'(0)=0

asked by guest
on Mar 26, 2025 at 5:35 pm



You asked:

Solve the initial value problem:y′′+5⋅y′+2y=e−tcos⁡(2t)y(0)=6y′(0)=0\begin{aligned}y^{\prime \prime} + 5 \cdot y^{\prime} + 2 y &= {e}^{-t} \cos\left( 2 t \right)\\y\left( 0 \right) &= 6\\y'\left( 0 \right) &= 0\end{aligned}

MathBot Answer:

y(t)=(7324+35917408)et(−5+17)2+(7324−35917408)e−t(17+5)2+e−tsin⁡(2t)12−e−tcos⁡(2t)12y{\left(t \right)} = \left(\frac{73}{24} + \frac{359 \sqrt{17}}{408}\right) e^{\frac{t \left(-5 + \sqrt{17}\right)}{2}} + \left(\frac{73}{24} - \frac{359 \sqrt{17}}{408}\right) e^{- \frac{t \left(\sqrt{17} + 5\right)}{2}} + \frac{e^{- t} \sin{\left(2 t \right)}}{12} - \frac{e^{- t} \cos{\left(2 t \right)}}{12}

ee is Euler's number, a mathematical constant that is approximately 2.718282.71828.

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