x''(t)-4x'(t)+4x(t)=4e^2t; x(0)=-1, x'(0)=-4

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asked by guest
on Mar 29, 2025 at 7:20 am



You asked:

Solve the initial value problem:x′′(t)−4⋅x′(t)+4⋅x(t)=4e2tx(0)=−1x′(0)=−4\begin{aligned}x''\left( t \right) - 4 \cdot x'\left( t \right) + 4 \cdot x\left( t \right) &= 4 {e}^{2} t\\x\left( 0 \right) &= -1\\x'\left( 0 \right) &= -4\end{aligned}

MathBot Answer:

x(t)=te2+(t(−2+e2)−e2−1)e2t+e2x{\left(t \right)} = t e^{2} + \left(t \left(-2 + e^{2}\right) - e^{2} - 1\right) e^{2 t} + e^{2}

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

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