solve y''+4y'+4y=(6+18t)e^-2t

and y(0)=2,y'(0)=0

asked by guest
on Nov 26, 2024 at 12:39 pm



You asked:

Solve the initial value problem:y′′+4⋅y′+4y=(6+18t)⋅e−2ty(0)=2y′(0)=0\begin{aligned}y^{\prime \prime} + 4 \cdot y^{\prime} + 4 y &= \left(6 + 18 t\right) \cdot {e}^{-2} t\\y\left( 0 \right) &= 2\\y'\left( 0 \right) &= 0\end{aligned}

MathBot Answer:

y(t)=9t22e2−15t2e2+(t(−3+4e2)e2+−21+8e24e2)e−2t+214e2y{\left(t \right)} = \frac{9 t^{2}}{2 e^{2}} - \frac{15 t}{2 e^{2}} + \left(\frac{t \left(-3 + 4 e^{2}\right)}{e^{2}} + \frac{-21 + 8 e^{2}}{4 e^{2}}\right) e^{- 2 t} + \frac{21}{4 e^{2}}

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

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