solve y''+4y'+4y=(6+18t)e^-2t
and y(0)=2,y'(0)=0
You asked:
Solve the initial value problem:\[\begin{aligned}y^{\prime \prime} + 4 \cdot y^{\prime} + 4 y &= \left(6 + 18 t\right) \cdot {e}^{-2} t\\y\left( 0 \right) &= 2\\y'\left( 0 \right) &= 0\end{aligned}\]
MathBot Answer:
\[y{\left(t \right)} = \frac{9 t^{2}}{2 e^{2}} - \frac{15 t}{2 e^{2}} + \left(\frac{t \left(-3 + 4 e^{2}\right)}{e^{2}} + \frac{-21 + 8 e^{2}}{4 e^{2}}\right) e^{- 2 t} + \frac{21}{4 e^{2}}\]
\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).