The Laplace transform of t2cosat is equal to L{t2 cosat}= 2s(s2-3a2)/(s2+a2)3. In this post, we will learn how to find the Laplace of t square cosat. The Laplace formula of t2cosat is given by $\boxed{ L\{t^2 \cos at \}= \dfrac{2s(s^2-3a^2)}{(s^2+a^2)^3} }$. What is the Laplace of t2 cosat Answer: The Laplace of t2 cosat is … Read more
The Laplace transform of cost cos2t is equal to s(s2+5)/(s2+1)(s2+9). Here we learn how to find the Laplace of cost cos2t. The Laplace transform formula of cost cos2t is given below: $\boxed{\mathcal{L} \{\cos t \cos 2t \} = \dfrac{s(s^2+5)}{(s^2+1)(s^2+9)} }$ Laplace of cost cos2t Answer: The Laplace of cost cos2t is s(s2+5) / (s4+10s2+9). Proof: … Read more
The Laplace transform of t^2, i.e the Laplace of t square is equal to 2/s3. In this article, we will learn how to find the Laplace transform of t2. The Laplace transform t2 (t square) is denoted by L{t2}, and its formula is given as follows: L{t2} = 2/s3. This follows from the Laplace formula … Read more
The Laplace transform of cos^4t is equal to 3/(8s) + s/[2(s2+4)] + s/[8(s2+16)]. So the Laplace formula of cos4t, that is, L{cos4t} is given as follows. L{cos4t} = $\dfrac{3}{8s} + \dfrac{s}{2s^2+8} + \dfrac{s}{8s^2+128}$. Let us now prove the above Laplace transform formula of the fourth power of cost. What is the Laplace of cos4t Answer: … Read more
The Laplace transform of sin^5t is equal to 5/(8s2+8) – 15/(16s2+144) + 5/(16s2+400). So the Laplace formula of sin5t is given as follows. L{sin5t} = $\dfrac{5}{8s^2+8} – \dfrac{15}{16s^2+144} + \dfrac{5}{16s^2+400}$. We now find the Laplace transform of the fifth power of sint. What is the Laplace of sin5t Answer: L{sin5t} = 5/(8s2+8) – 15/(16s2+144) + … Read more
The Laplace transform of sin^4t is denoted by L{sin4t}, and it is equal to 3/(8s) – s/[2(s2+4)] + s/[8(s2+16)]. So the Laplace formula of sin4t is equal to L{sin4t} = $\dfrac{3}{8s} – \dfrac{s}{2(s^2+4)} + \dfrac{s}{8(s^2+16)}$. Let us now prove the above Laplace transform formula of the fourth power of sint. Laplace of sin4t Question: What … Read more
The Laplace transform formula of periodic functions is used to find the Laplace of a periodic function with period T, that is, f(t+T)=f(t). This formula says that the Laplace transform of f(t) is given by L{f(t)} = $\dfrac{\int_0^T e^{-st} f(t) dt}{1-e^{-sT}}$. Laplace Transform of a Periodic Function Theorem: If f: [0, ∞) → ℝ is a … Read more
The Laplace transform of t^2 u(t-1) is equal to e-s[2/s3 + 2/s2 +1/s]. Here we find the Laplace of t2u(t-1) using the second shifting property of Laplace transforms. The formula of the Laplace of t2u(t-1) is given as L{t2u(t-1)} = $e^{-s}\Big[ \dfrac{2}{s^3} + \dfrac{2}{s^2} +\dfrac{1}{s}\Big]$. What is the Laplace of t2u(t-1)? Answer: The Laplace of … Read more
The Laplace transform of u(t-2) is equal to e-2s/s and it is denoted by L{u(t-2)} = e-2s/s. Before we find the Laplace of u(t-2), the shifted unit step function by 2, let us first recall the definition of u(t-2): $u(t-2)= \begin{cases} 0 & \text{ if } t<2 \\ 1 & \text{ if } t \geq … Read more
The Laplace transform of u(t-1) is equal to e-s/s, that is, L{u(t-1)} = e-s/s. Note that u(t-1) is the shifted unit step function by 1 and it is defined as follows. u(t-1) = 0 if t<1 u(t-1) = 1 if t≥1. Let us now learn how to find the Laplace transform of u(t-1). Laplace of … Read more