Take the Laplace transform of the initial value...

Take the Laplace transform of the initial value problem\\ $\frac{d^2y}{dt^2} + k^2y = e^{-2t}$, $y(0) = 0$, $y'(0) = 0$.\\ $\boxed{} = \frac{1}{s+2}$ help (formulas)\\ Note: Enter the equation as it drops out of the Laplace transform, do not move terms from one side to the other yet. Use Y for the Laplace transform of y(t), (not\\$Y(s)$)\ So\\$Y = \frac{1}{(s+2)(s^2 + k^2)}$\\$= \frac{\boxed{}}{s+2} + \frac{\boxed{s} + \boxed{}}{s^2 + k^2}$\\help (formulas)\\and\\$y(t) = \boxed{}$ help (formulas)

Transcript

00:01 These students here we have a question and we have to use laplace transformation to solve this initial value problem so these students the initial conditions are y of 0 is equal to 0 and y prime of 0 is equal to 1 so these students let's apply laplace on each term it will be a plus plus y double prime plus 2 laplace of y prime is equal to le plus of le plus of le plus of delta into t minus 1 so these students notice that here we have laplace y double prime is equal to s square y of s minus s y of 0 minus y prime of 0 and laplace y prime is equal to s y of s minus y of 0 and dear student notice that we have the initial conditions y of 0 is equal to 0 and y prime of 0 is equal to 1 so de s students now if you put the initial conditions right here the formula will give us this will give us s square y of s and this will be this will be equal to de students minus one plus two s y of s is equal to and de students we know that the laplace of this the general form laplace and laplace of delta into t minus a is equal to e to the power minus a s and here we have a in our perspective it is one so this will give us e to the power minus so now the students if you take y of s common and put if you take s y of s common if you take s y of s common it will give us s plus 2 into s plus 2 into now s of y of s plus 2 is equal to e to the power minus s and we put minus 1 into the right -hand side it will give plus 1 give plus 1 give plus so now the student y of s will be equal to e to the power minus s divided by s into s plus plus one divided by s into s plus two so now these students if we apply le plus inverse it will it will give us the plus inverse of y of s is equal to the plus inverse of e to the power minus s divided by s into s plus two plus le plus inverse 1 divided by s into s plus 2 so these students here notice that this here notice that we have a formula we have a formula we have a theorem which is theorem tells us which is the inverse second translation theorem a plus inverse of e to the power minus a s f of s equals to what f into t minus a and mu into t minus a so now the students here in our perspective of s is what it is 1 divided by s into s plus 2 so this then if we solve it through s into s plus 1 if we solve it through partial fraction it will give a divided by s plus b divided by s plus 1 so this student if you put s is equal to 0 a will be equal to 1 divided by 2 and if we put s is equal to minus it is dea student it is here notice that it is here it is two this two and right here it is also it is also so this will give this will give if you put s is equal to minus 2 b will be equal to minus 1 upon 2 so these students now so now day…

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