Answer to Question #208299 in Differential Equations for Abuabu

Question

Find the laplace transform of sinat where a is a constant

Answer 1

$F(p) = \int\limits_0^\infty {f(t){e^{ - pt}}} dt = \int\limits_0^\infty {\sin at \cdot {e^{ - pt}}} dt = \left| {\begin{matrix} {u = \sin at}&{dv = {e^{ - pt}}dt}\\ {du = a\cos atdt}&{v = - \frac{1}{p}{e^{ - pt}}} \end{matrix}} \right| = \mathop {\lim }\limits_{x \to \infty } \left( { - \left. {\frac{1}{p}\sin at \cdot {e^{ - pt}}} \right|_0^x + \frac{a}{p}\int\limits_0^x {\cos at{e^{ - pt}}} dt} \right) = \left| {\begin{matrix} {u = \cos at}&{dv = {e^{ - pt}}dt}\\ {du = - a\sin atdt}&{v = - \frac{1}{p}{e^{ - pt}}} \end{matrix}} \right| = \mathop {\lim }\limits_{x \to \infty } \left( { - \left. {\frac{1}{p}\sin at \cdot {e^{ - pt}}} \right|_0^x + \frac{a}{p}\left( { - \frac{1}{p}\cos at \cdot \left. {{e^{ - pt}}} \right|_0^x - \frac{a}{p}\int\limits_0^x {\sin at \cdot {e^{ - pt}}} dt} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } \left( { - \left. {\frac{1}{p}\sin at \cdot {e^{ - pt}}} \right|_0^x - \frac{a}{{{p^2}}}\cos at \cdot \left. {{e^{ - pt}}} \right|_0^x - \frac{{{a^2}}}{{{p^2}}}\int\limits_0^x {\sin at \cdot {e^{ - pt}}} dt} \right) = 0 + 0 - 0 + \frac{a}{{{p^2}}}\cos 0 \cdot {e^0} - \frac{{{a^2}}}{{{p^2}}}\mathop {\lim }\limits_{x \to \infty } \int\limits_0^x {\sin at \cdot {e^{ - pt}}} dt = \frac{a}{{{p^2}}} - \frac{{{a^2}}}{{{p^2}}}\mathop {\lim }\limits_{x \to \infty } \int\limits_0^x {\sin at \cdot {e^{ - pt}}} dt$

Then

$F(p) = \frac{a}{{{p^2}}} - \frac{{{a^2}}}{{{p^2}}}F(p) \Rightarrow F(p)\left( {1 + \frac{{{a^2}}}{{{p^2}}}} \right) = \frac{a}{{{p^2}}} \Rightarrow F(p) \cdot \frac{{{p^2} + {a^2}}}{{{p^2}}} = \frac{a}{{{p^2}}} \Rightarrow F(p) = \frac{a}{{{p^2} + {a^2}}}$

Answer: $F(p) = \frac{a}{{{p^2} + {a^2}}}$