Question

Find Laplace transform X(p) for function (show all work) in differential equation:
x''''(t)-5x'''(t)-4x''(t)+2x'(t)-x(t)+8=0
x(0)=5
x'(0)-0
x''(0)=-1
x'''(0)=2

Answer 1

$$x ^ {\prime \prime \prime \prime} (t) - 5 x ^ {\prime \prime \prime} (t) - 4 x ^ {\prime \prime} (t) + 2 x ^ {\prime} (t) - x (t) + 8 = 0$$ $$x (t) \rightarrow X (p)$$

Then use differentiation theorem:

$$x ^ {\prime} (t) \rightarrow p X (p) - 5$$ $$x ^ {\prime \prime} (t) \rightarrow p ^ {2} X (p) - 5 p$$ $$x ^ {\prime \prime \prime} (t) \rightarrow p ^ {3} X (p) - 5 p ^ {2} + 1$$ $$x ^ {\prime \prime \prime \prime} (t) \rightarrow p ^ {4} X (p) - 5 p ^ {3} + p - 2$$

By liniarization:

$$\begin{array}{l} x ^ {\prime \prime \prime \prime} (t) - 5 x ^ {\prime \prime \prime} (t) - 4 x ^ {\prime \prime} (t) + 2 x ^ {\prime} (t) - x (t) + 8 \rightarrow \\ \rightarrow p ^ {4} X (p) - 5 p ^ {3} + p - 2 - 5 \left(p ^ {3} X (p) - 5 p ^ {2} + 1\right) - \\ - 4 \left(p ^ {2} X (p) - 5 p\right) + 2 \left(p X (p) - 5\right) - X (p) + \frac {8}{p} = \\ = \left(p ^ {4} - 5 p ^ {3} - 4 p ^ {2} + 2 p - 1\right) X (p) - 5 p ^ {3} + 2 5 p ^ {2} + 2 1 p - 1 7 + 8 / p \\ \end{array}$$