Compute the inverse Laplace transform of
$$\frac{3s +1}{{s^{3}}+{4s^{2}}+{(k-3)}}$$
Already tried with partial fraction expansion, without success.
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Why did partial fraction expansion fail? – Mark Viola Feb 07 '19 at 03:15
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@MarkViola As of my approach , the polynomial $${{s^{3}}+{4s^{2}}+{(k-3)}}$$ is not completely reducible by the numerator $${3s +1}$$ It has format $$D(s)=N(s)Q(s) + R(s)$$ , where D(s) is : $${{s^{3}}+{4s^{2}}+{(k-3)}}$$ , N(s) = $${3s +1}$$ , Q(s) is the quotient polynomial and R(s) is the remainder polynomial – Nilabja Saha Feb 07 '19 at 03:20
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See THIS. – Mark Viola Feb 07 '19 at 03:49
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Regardless of the coefficients, the roots of any cubic polynomial can be found in closed form. The location of the roots obviously depend on $k$. – Mark Viola Feb 07 '19 at 04:11