The laplace transform of sin 2t sin 3t is

Answers 1

[tex]\large\underline{\sf{Solution-}}[/tex]

Given expression is

[tex]\rm \: L( \: sin2t \: sin3t \: ) \\ \\ [/tex]

can be rewritten as

[tex]\rm \: = \: \dfrac{1}{2} L( \: 2 \: sin2t \: sin3t \: ) \\ \\ [/tex]

We know,

[tex]\boxed{ \rm{ \:2 \: sinx \: siny \: = \: cos(x - y) \: - \: cos(x + y) \: }} \\ \\ [/tex]

So, using this result, we get

[tex]\rm \: = \: \dfrac{1}{2} \: L\bigg[\: cos(2t - 3t) - cos(3t + 2t) \: \bigg] \\ \\ [/tex]

[tex]\rm \: = \: \dfrac{1}{2} \: L\bigg[\: cos( - t) - cos(5t) \: \bigg] \\ \\ [/tex]

[tex]\rm \: = \: \dfrac{1}{2} \: L\bigg[\: cos(t) - cos(5t) \: \bigg] \\ \\ [/tex]

We know,

[tex]\boxed{ \rm{ \:L(cosat) \: = \: \dfrac{s}{ {s}^{2} + {a}^{2} } \: }} \\ \\ [/tex]

So, using this result, we get

[tex]\rm \: = \: \dfrac{1}{2} \bigg(\dfrac{s}{ {s}^{2} + {1}^{2} } - \dfrac{s}{ {s}^{2} + {5}^{2} } \bigg) \\ \\ [/tex]

[tex]\rm \: = \: \dfrac{1}{2} \bigg(\dfrac{s}{ {s}^{2}+1} - \dfrac{s}{ {s}^{2}+ 25} \bigg) \\ \\ [/tex]

Hence,

[tex]\rm\implies \boxed{ \rm{L( \: sin2t \: sin3t \: )=\dfrac{1}{2} \bigg(\dfrac{s}{ {s}^{2}+1}-\dfrac{s}{ {s}^{2}+ 25} \bigg) \: }} \\ \\ [/tex]

[tex]\rule{190pt}{2pt} \\ [/tex]

[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]

[tex]\boxed{ \rm{ \:L( {e}^{at}) = \frac{1}{s - a} \: }} \\ [/tex]

[tex]\boxed{ \rm{ \:L( {e}^{ - at}) = \frac{1}{s + a} \: }} \\ [/tex]

[tex]\boxed{ \rm{ \:L( sinat) = \frac{a}{ {s}^{2} + {a}^{2} } \: }} \\ [/tex]

[tex]\boxed{ \rm{ \:L( {e}^{bt} sinat) = \frac{a}{ {(s - b)}^{2} + {a}^{2} } \: }} \\ [/tex]

[tex]\boxed{ \rm{ \:L( {e}^{bt} cosat) \: = \: \dfrac{s}{ {(s - b)}^{2} + {a}^{2} } \: }} \\ \\ [/tex]